login
A062786
Centered 10-gonal numbers.
40
1, 11, 31, 61, 101, 151, 211, 281, 361, 451, 551, 661, 781, 911, 1051, 1201, 1361, 1531, 1711, 1901, 2101, 2311, 2531, 2761, 3001, 3251, 3511, 3781, 4061, 4351, 4651, 4961, 5281, 5611, 5951, 6301, 6661, 7031, 7411, 7801, 8201, 8611, 9031, 9461, 9901, 10351, 10811
OFFSET
1,2
COMMENTS
Deleting the least significant digit yields the (n-1)-st triangular number: a(n) = 10*A000217(n-1) + 1. - Amarnath Murthy, Dec 11 2003
All divisors of a(n) are congruent to 1 or -1, modulo 10; that is, they end in the decimal digit 1 or 9. Proof: If p is an odd prime different from 5 then 5n^2 - 5n + 1 == 0 (mod p) implies 25(2n - 1)^2 == 5 (mod p), whence p == 1 or -1 (mod 10). - Nick Hobson, Nov 13 2006
Centered decagonal numbers. - Omar E. Pol, Oct 03 2011
The partial sums of this sequence give A004466. - Leo Tavares, Oct 04 2021
The continued fraction expansion of sqrt(5*a(n)) is [5n-3; {2, 2n-2, 2, 10n-6}]. For n=1, this collapses to [2; {4}]. - Magus K. Chu, Sep 12 2022
Numbers m such that 20*m + 5 is a square. Also values of the Fibonacci polynomial y^2 - x*y - x^2 for x = n and y = 3*n - 1. This is a subsequence of A089270. - Klaus Purath, Oct 30 2022
All terms can be written as a difference of two consecutive squares a(n) = A005891(n-1)^2 - A028895(n-1)^2, and they can be represented by the forms (x^2 + 2mxy + (m^2-1)y^2) and (3x^2 + (6m-2)xy + (3m^2-2m)y^2), both of discriminant 4. - Klaus Purath, Oct 17 2023
FORMULA
a(n) = 5*n*(n-1) + 1.
From Gary W. Adamson, Dec 29 2007: (Start)
Binomial transform of [1, 10, 10, 0, 0, 0, ...];
Narayana transform (A001263) of [1, 10, 0, 0, 0, ...]. (End)
a(n) = 10*n + a(n-1) - 10; a(1)=1. - Vincenzo Librandi, Aug 07 2010
G.f.: x*(1+8*x+x^2) / (1-x)^3. - R. J. Mathar, Feb 04 2011
a(n) = A124080(n-1) + 1. - Omar E. Pol, Oct 03 2011
a(n) = A101321(10,n-1). - R. J. Mathar, Jul 28 2016
a(n) = A028387(A016861(n-1))/5 for n > 0. - Art Baker, Mar 28 2019
E.g.f.: (1+5*x^2)*exp(x) - 1. - G. C. Greubel, Mar 30 2019
Sum_{n>=1} 1/a(n) = Pi * tan(Pi/(2*sqrt(5))) / sqrt(5). - Vaclav Kotesovec, Jul 23 2019
From Amiram Eldar, Jun 20 2020: (Start)
Sum_{n>=1} a(n)/n! = 6*e - 1.
Sum_{n>=1} (-1)^n * a(n)/n! = 6/e - 1. (End)
a(n) = A005891(n-1) + 5*A000217(n-1). - Leo Tavares, Jul 14 2021
a(n) = A003154(n) - 2*A000217(n-1). See Mid-section Stars illustration. - Leo Tavares, Sep 06 2021
From Leo Tavares, Oct 06 2021: (Start)
a(n) = A144390(n-1) + 2*A028387(n-1). See Mid-section Star Pillars illustration.
a(n) = A000326(n) + A000217(n) + 3*A000217(n-1). See Trapezoidal Rays illustration.
a(n) = A060544(n) + A000217(n-1). (End)
From Leo Tavares, Oct 31 2021: (Start)
a(n) = A016754(n-1) + 2*A000217(n-1).
a(n) = A016754(n-1) + A002378(n-1).
a(n) = A069099(n) + 3*A000217(n-1).
a(n) = A069099(n) + A045943(n-1).
a(n) = A003215(n-1) + 4*A000217(n-1).
a(n) = A003215(n-1) + A046092(n-1).
a(n) = A001844(n-1) + 6*A000217(n-1).
a(n) = A001844(n-1) + A028896(n-1).
a(n) = A005448(n) + 7*A000217(n).
a(n) = A005448(n) + A024966(n). (End)
From Klaus Purath, Oct 30 2022: (Start)
a(n) = a(n-2) + 10*(2*n-3).
a(n) = 2*a(n-1) - a(n-2) + 10.
a(n) = A135705(n-1) + n.
a(n) = A190816(n) - n.
a(n) = 2*A005891(n-1) - 1. (End)
MATHEMATICA
FoldList[#1+#2 &, 1, 10Range@ 45] (* Robert G. Wilson v, Feb 02 2011 *)
1+5*Pochhammer[Range[50]-1, 2] (* G. C. Greubel, Mar 30 2019 *)
PROG
(PARI) j=[]; for(n=1, 75, j=concat(j, (5*n*(n-1)+1))); j
(PARI) for (n=1, 1000, write("b062786.txt", n, " ", 5*n*(n - 1) + 1) ) \\ Harry J. Smith, Aug 11 2009
(Magma) [1+5*n*(n-1): n in [1..50]]; // G. C. Greubel, Mar 30 2019
(Sage) [1+5*rising_factorial(n-1, 2) for n in (1..50)] # G. C. Greubel, Mar 30 2019
(GAP) List([1..50], n-> 1+5*n*(n-1)) # G. C. Greubel, Mar 30 2019
(Python) def a(n): return(5*n**2-5*n+1) # Torlach Rush, May 10 2024
KEYWORD
easy,nonn
AUTHOR
Jason Earls, Jul 19 2001
EXTENSIONS
Better description from Terrel Trotter, Jr., Apr 06 2002
STATUS
approved