A168362 a(n) = coefficient of x^n in the n-th iteration of Sum_{k>=0} x^(2^k), n>=1.

1, 2, 6, 34, 280, 3010, 39984, 634040, 11704548, 246799212, 5856139256, 154509170816, 4488522398568, 142395677932872, 4899139202191216, 181714457372434436, 7228856213182113768, 307047663830976178304, 13869994589640136690336, 663976989978171594869350, 33579260181092227231600048

Offset

1,2

Comments

It appears that a(n+1)/a(n) ~ n*exp(1). - Paul D. Hanna, May 22 2026

Links

Paul D. Hanna, Table of n, a(n) for n = 1..520

Example

Given F(x) = x + x^2 + x^4 + x^8 + x^16 +...+ x^(2^n) + ...,
define F^n(x) = F^(n-1)(F(x)) as the n-th iteration of F(x);
then the table of coefficients of x^k in F^n(x) begins
  n = 1: [1, 1,  0,   1,    0,     0,     0,      1, ...];
  n = 2: [1, 2,  2,   3,    6,     8,     8,     16, ...];
  n = 3: [1, 3,  6,  12,   28,    68,   152,    330, ...];
  n = 4: [1, 4, 12,  34,  100,   310,   972,   3012, ...];
  n = 5: [1, 5, 20,  75,  280,  1070,  4176,  16449, ...];
  n = 6: [1, 6, 30, 141,  650,  3010, 14116,  66972, ...];
  n = 7: [1, 7, 42, 238, 1316,  7238, 39984, 222614, ...];
  n = 8: [1, 8, 56, 372, 2408, 15428, 98728, 634040, ...]; ...
in which the main diagonal forms the initial terms of this sequence.
For n > 1, a(n) == 2 (mod 4) when n = [2, 3, 4, 6, 20, 24, 40, 68, 136, 260, 520, ...], and it appears that a(n) == 0 (mod 4) elsewhere. - Paul D. Hanna, May 22 2026

Prog

(PARI) {a(n) = my(A=vector(n), G=x, B = sum(k=0, #binary(n), x^(2^k)) +x*O(x^n)); A[#A]=1; for(m=1, n, G=subst(B, x, G); A[m] = polcoeff(G, m)); AV=A; A[n]}
{upto(n) = a(n); AV}
upto(20) \\ program revised by Paul D. Hanna, May 22 2026

Crossrefs

Sequence in context: A271212 A325296 A052878 * A274711 A076863 A395864

Adjacent sequences: A168359 A168360 A168361 * A168363 A168364 A168365

Keyword

nonn

Author

Paul D. Hanna, Nov 23 2009

Extensions

Entry revised by Paul D. Hanna, May 22 2026

Status

approved