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A247489
Square array read by antidiagonals: A(k, n) = hypergeometric(P, Q, -k^k/(k-1)^(k-1)) rounded to the nearest integer, P = [(j-n)/k, j=0..k-1] and Q = [(j-n)/(k-1), j=0..k-2], k>=1, n>=0.
1
1, 0, 2, 0, 1, 4, 0, 1, 2, 8, 0, 1, 1, 3, 16, 0, 1, 1, 2, 5, 32, 0, 1, 1, 1, 3, 8, 64, 0, 0, 1, 1, 2, 4, 13, 128, 0, 0, 1, 1, 1, 3, 6, 21, 256, 0, 0, 1, 1, 1, 2, 4, 9, 34, 512, 0, 0, 1, 1, 1, 1, 3, 5, 13, 55, 1024, 0, 0, 1, 1, 1, 1, 2, 4, 7, 19, 89, 2048
OFFSET
0,3
COMMENTS
Conjecture: hypergeometric(P, Q, -k^k/(k-1)^(k-1)) = sum_{j=0.. floor(n/k)} binomial(n-(k-1)*j, j) for n>=(k-1)^2, P and Q as above. (This means for n>=(k-1)^2 the representation is exact without rounding.)
EXAMPLE
First few rows of the square array:
[k\n] if conjecture true
[1], 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, ... A000079 n>=0
[2], 0, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ... 'A000045' n>=1
[3], 0, 1, 1, 2, 3, 4, 6, 9, 13, 19, 28, 41, ... A000930 n>=4
[4], 0, 1, 1, 1, 2, 3, 4, 5, 7, 10, 14, 19, ... A003269 n>=9
[5], 0, 1, 1, 1, 1, 2, 3, 4, 5, 6, 9, 11, 15, ... A003520 n>=16
[6], 0, 1, 1, 1, 1, 1, 2, 3, 3, 4, 6, 7, 10, ... A005708 n>=25
[7], 0, 0, 1, 1, 1, 1, 1, 2, 3, 3, 4, 5, 7, 8, ... A005709 n>=36
[8], 0, 0, 1, 1, 1, 1, 2, 1, 2, 3, 3, 4, 5, 6, ... A005710 n>=49
'A000045' means that the Fibonacci numbers as referenced here start 1, 1, 2, 3, ... for n>=0.
MAPLE
A247489 := proc(k, n)
seq((j-n)/k, j=0..k-1); seq((j-n)/(k-1), j=0..k-2);
hypergeom([%%], [%], -k^k/(k-1)^(k-1));
round(evalf(%, 100)) end: # Adjust precision if necessary!
for k from 1 to 9 do print(seq(A247489(k, n), n=0..16)) od;
PROG
(Sage)
def A247489(k, n):
P = [(j-n)/k for j in range(k)]
Q = [(j-n)/(k-1) for j in range(k-1)]
H = hypergeometric(P, Q, -k^k/(k-1)^(k-1))
return round(H.n(100)) # Adjust precision if necessary!
KEYWORD
nonn,tabl
AUTHOR
Peter Luschny, Sep 19 2014
STATUS
approved