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Suppose that A = (a(n)), for n >= 0, is a sequence, and g is a real number such that a(n)/a(n-1) -> g. The ratio-sum for A is |a(1)/a(0) - g| + |a(2)/a(1) - g| + . . . , , assuming that this series converges. For A = A295952, we have g = (1 + sqrt(5))/2, the golden ratio (A001622). See the guide at A296469 for related sequences.

While[k < a[j + 1] - j + 1, b[k] = j + k + 2; k++]; j++];

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allocated Decimal expansion of ratio-sum for Clark KimberlingA295952; see Comments.

4, 8, 4, 5, 8, 5, 3, 6, 8, 3, 5, 1, 4, 3, 6, 2, 0, 7, 1, 3, 5, 0, 0, 2, 0, 5, 6, 7, 3, 7, 2, 5, 0, 1, 7, 8, 9, 0, 3, 4, 8, 4, 3, 5, 6, 2, 3, 5, 7, 9, 0, 5, 1, 6, 3, 2, 0, 5, 9, 9, 3, 0, 5, 7, 2, 8, 9, 5, 5, 2, 9, 0, 7, 4, 0, 0, 5, 7, 1, 0, 7, 9, 6, 9, 5, 0

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Suppose that A = (a(n)), for n >= 0, is a sequence, and g is a real number such that a(n)/a(n-1) -> g. The ratio-sum for A is |a(1)/a(0) - g| + |a(2)/a(1) - g| + . . . , assuming that this series converges. For A = A295952, we have g = (1 + sqrt(5))/2, the golden ratio (A001622). See the guide at A296469 for related sequences.

ratio-sum = 4.845853683514362071350020567372501789034...

a[0] = 1; a[1] = 5; b[0] = 2; b[1 ] = 3; b[2] = 4;

a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n];

j = 1; While[j < 13, k = a[j] - j - 1;

While[k < a[j + 1] - j + 1, b[k] = j + k + 2; k++]; j++];

Table[a[n], {n, 0, k}]; (* A295952 *)

g = GoldenRatio; s = N[Sum[- g + a[n]/a[n - 1], {n, 1, 1000}], 200]

Take[RealDigits[s, 10][[1]], 100] (* A296481 *)

Cf. A001622, A296284, A296482.

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nonn,easy,cons

Clark Kimberling, Jan 05 2018

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