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 A333096 a(n) = the n-th order Taylor polynomial (centered at 0) of c(x)^(4*n) evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. of the sequence of Catalan numbers A000108. 4
 1, 5, 53, 647, 8373, 111880, 1525511, 21093476, 294663349, 4148593604, 58770091928, 836722722951, 11961868391175, 171601856667701, 2469036254872996, 35615467194043147, 514888180699419829, 7458193213805231529, 108219144962546395364, 1572690742149983040857 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS The sequence satisfies the Gauss congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^k ) for all prime p and positive integers n and k. We conjecture that the sequence satisfies the stronger congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k. Examples of these congruences are given below. More generally, for each integer m, we conjecture that the sequence a_m(n) := the n-th order Taylor polynomial of c(x)^(m*n) evaluated at x = 1 satisfies the same congruences. For cases see A099837 (m = -2), A100219 (m = -1), A000012 (m = 0), A333093 (m = 1), A333094 (m = 2), A333095 (m = 3), A333097 (m = 5). LINKS FORMULA a(n) = Sum_{k = 0..n} 4*n/(4*n+k)*binomial(4*n+2*k-1,k) for n >= 1. a(n) = [x^n] ( (1 + x)*c^4(x/(1 + x)) )^n. O.g.f.: ( 1 + x*f'(x)/f(x) )/( 1 - x*f(x) ), where f(x) = 1 + 4*x + 30*x^2 + 280*x^3 + 2925*x^4 + ... = (1/x)*Revert( x/c^4(x) ) is the o.g.f. of A212073. Row sums of the Riordan array ( 1 + x*f'(x)/f(x), f(x) ) belonging to the Hitting time subgroup of the Riordan group. a(n) ~ 2^(6*n + 3) * 3^(6*n + 3/2) / (31 * sqrt(Pi*n) * 5^(5*n + 1/2)). - Vaclav Kotesovec, Mar 28 2020 EXAMPLE n-th order Taylor polynomial of c(x)^(4*n):   n = 0: c(x)^0 = 1 + O(x)   n = 1: c(x)^4 = 1 + 4*x + O(x^2)   n = 2: c(x)^8 = 1 + 8*x + 44*x^2 + O(x^3)   n = 3: c(x)^12 = 1 + 12*x + 90*x^2 + 544*x^3 + O(x^4)   n = 4: c(x)^16 = 1 + 16*x + 152*x^2 + 1120*x^3 + 7084*x^4 + O(x^5) Setting x = 1 gives a(0) = 1, a(1) = 1 + 4 = 5, a(2) = 1 + 8 + 44 = 53, a(3) = 1 + 12 + 90 + 544 = 647 and a(4) = 1 + 16 + 152 + 1120 + 7084 = 8373. The triangle of coefficients of the n-th order Taylor polynomial of c(x)^(4*n), n >= 0, in descending powers of x begins                                          row sums   n = 0 |    1                               1   n = 1 |    4     1                         5   n = 2 |   44     8     1                  53   n = 3 |  544    90    12     1           647   n = 4 | 7084  1120   152    16   1      8373    ... This is a Riordan array belonging to the Hitting time subgroup of the Riordan group. Examples of congruences: a(13) - a(1) = 171601856667701 - 5 = (2^4)*3*(7^2)*(13^3)*33208909 == 0 ( mod 13^3 ). a(3*7) - a(3) = 333475516822140871773101 - 647 = 2*(3^2)*(7^3)* 54012879303877692221 == 0 ( mod 7^3 ). a(5^2) - a(5) = 15187725485911657497382846255 - 111880 = (3^3)*(5^7)*29* 248279548173268475053 == 0 ( mod 5^6 ). MAPLE seq(add(4*n/(4*n+k)*binomial(4*n+2*k-1, k), k = 0..n), n = 1..25); #alternative program c:= x → (1/2)*(1-sqrt(1-4*x))/x: G := (x, n) → series(c(x)^(4*n), x, 126): seq(add(coeff(G(x, n), x, n-k), k = 0..n), n = 0..25); MATHEMATICA Join[{1}, Table[4*Binomial[6*n-1, n] * HypergeometricPFQ[{1, -5*n, -n}, {1/2 - 3*n, 1 - 3*n}, 1/4]/5, {n, 1, 20}]] (* Vaclav Kotesovec, Mar 28 2020 *) CROSSREFS Cf. A000108, A212073, A333090 through A333097. Sequence in context: A065534 A186206 A123788 * A036910 A235371 A036916 Adjacent sequences:  A333093 A333094 A333095 * A333097 A333098 A333099 KEYWORD nonn,easy AUTHOR Peter Bala, Mar 15 2020 STATUS approved

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Last modified May 27 13:56 EDT 2022. Contains 354097 sequences. (Running on oeis4.)