OFFSET
1,1
COMMENTS
There exists one exceptional case of 4 consecutive numbers 12, 13, 14, 15, where 13 would start an odd-even-odd progression.
LINKS
Hugo Pfoertner, Table of n, a(n) for n = 1..1000
Max Alekseyev, PARI/GP Scripts for Miscellaneous Math Problems: invphi.gp, Oct. 2005.
Jeppe Stig Nielsen, List of numbers with divisor sum k, k<=10000.
EXAMPLE
a(1) = 3, because 2*3 = 6 is the start of the first occurrence of a row of 3 consecutive numbers, all of which are in A000203. 6 = sigma(5), 7 = sigma(4), 8 = sigma(7).
a(2) = 6: 2*6 = 12 = sigma(6) = sigma(11), 13 = sigma(9), 14 = sigma(13). 15 = sigma(8), which would be at the end of the row 13, 14, 15, is excluded by the even-odd-even condition.
a(3) = 15: 2*15 = 30 = sigma(29), 31 = sigma(16) = sigma(25), 32 = sigma(21) = sigma(31).
See Jeppe Stig Nielsen's list for more examples.
PROG
(PARI) a342555(nterms) = {my(N=vector(3, i, invsigmaNum(i+1)), n=0, k=4); while(n<=nterms, if(vecmin(N)>0 && !(k%2), print1((k-2)/2, ", "); n++); k++; N[1+k%3] = invsigmaNum(k))}; \\ see Alekseyev link for invsigmaNum()
a342555(46)
CROSSREFS
KEYWORD
nonn
AUTHOR
Hugo Pfoertner, May 14 2021
STATUS
approved