A386654 a(n) = Sum_{k>=0} ( ( binomial(n+k-1, k) * 4^k ) (mod 5^k) ) / 5^k.

4, 11, 18, 28, 35, 45, 48, 60, 68, 78, 88, 87, 97, 113, 120, 129, 138, 142, 147, 165, 174, 179, 183, 196, 200, 223, 236, 235, 242, 252, 258, 266, 266, 291, 283, 290, 305, 314, 335, 337, 344, 355, 348, 370, 378, 395, 391, 407, 424, 417, 417, 432, 461, 460, 464, 471, 500, 481, 506, 501, 523, 525, 527, 527, 545, 555, 572, 575

Offset

1,1

Links

Paul D. Hanna, Table of n, a(n) for n = 1..500

Example

The terms a(n) equal the sum as illustrated below.
a(1) = (1 mod 1) + (4 mod 5)/5 + (16 mod 5^2)/5^2 + (64 mod 5^3)/5^3 + ...
a(2) = (1 mod 1) + (8 mod 5)/5 + (48 mod 5^2)/5^2 + (256 mod 5^3)/5^3 + ...
a(3) = (1 mod 1) + (12 mod 5)/5 + (96 mod 5^2)/5^2 + (640 mod 5^3)/5^3 + ...
a(4) = (1 mod 1) + (16 mod 5)/5 + (160 mod 5^2)/5^2 + (1280 mod 5^3)/5^3 + ...
a(5) = (1 mod 1) + (20 mod 5)/5 + (240 mod 5^2)/5^2 + (2240 mod 5^3)/5^3 + ...
...
More explicitly,
a(1) = 0 + 4/5 + 16/5^2 + 64/5^3 + 256/5^4 + 1024/5^5 + ...
a(2) = 0 + 3/5 + 23/5^2 + 6/5^3 + 30/5^4 + 3019/5^5 + ...
a(3) = 0 + 2/5 + 21/5^2 + 15/5^3 + 90/5^4 + 2754/5^5 + ...
a(4) = 0 + 1/5 + 10/5^2 + 30/5^3 + 210/5^4 + 1094/5^5 + ...
a(5) = 0 + 0/5 + 15/5^2 + 115/5^3 + 420/5^4 + 899/5^5 + ...
a(6) = 0 + 4/5 + 11/5^2 + 84/5^3 + 381/5^4 + 1798/5^5 + ...
a(7) = 0 + 3/5 + 23/5^2 + 1/5^3 + 10/5^4 + 1213/5^5 + ...
a(8) = 0 + 2/5 + 1/5^2 + 55/5^3 + 105/5^4 + 1633/5^5 + ...
...

Prog

(PARI) {a(n) = round( sum(k=0, 25*n, ( (binomial(n+k-1, k) * 4^k) % 5^k) / 5^k *1. ) )}
for(n=1, 60, print1(a(n), ", "))

Crossrefs

Cf. A386651, A386652, A386653.

Sequence in context: A339215 A242679 A017029 * A009873 A300744 A300743

Adjacent sequences: A386651 A386652 A386653 * A386655 A386656 A386657

Keyword

nonn

Author

Paul D. Hanna, Jul 27 2025

Status

approved