A059737
Lesser of the smallest pair of consecutive numbers divisible by an n-th power, but neither divisible by an (n+1)-st power.
Original entry on oeis.org
2, 44, 135, 80, 8991, 29888, 356480, 2316032, 14073344, 24151040, 326481920, 689278976, 11573190656, 76876660736, 314944159743, 2035980763136, 28996228218879, 55637069004800, 766556765683712, 1375916505694208, 19656708706009088, 129341461907898368, 2280241934368767, 787449981119234048
Offset: 0
Original entry on oeis.org
1, 25, 362, 3973, 36646, 299530, 2238676, 15613741, 103054094, 650194974, 3950996556, 23257207714, 133217073276, 745218012084, 4083224828328, 21966983072637, 116268166691358, 606474982072982, 3122157367765788
Offset: 1
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R:=PowerSeriesRing(Rationals(), 40); Coefficients(R!( (1-Sqrt(1-4*x))/(2*(1-4*x)^6) )); // G. C. Greubel, Jan 13 2020
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seq(coeff(series((1-sqrt(1-4*x))/(2*(1-4*x)^6), x, n+1), x, n), n = 0..40); # G. C. Greubel, Jan 13 2020
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CoefficientList[Series[(1-Sqrt[1-4*x])/(2*x*(1-4*x)^6), {n,0,40}], x] (* G. C. Greubel, Jan 13 2020 *)
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my(x='x+O('x^40)); Vec((1-sqrt(1-4*x))/(2*(1-4*x)^6)) \\ G. C. Greubel, Jan 13 2020
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def A045622_list(prec):
P. = PowerSeriesRing(ZZ, prec)
return P( (1-sqrt(1-4*x))/(2*(1-4*x)^6) ).list()
A045622_list(40) # G. C. Greubel, Jan 13 2020
Original entry on oeis.org
1, 3, 1, 10, 7, 1, 35, 38, 11, 1, 126, 187, 82, 15, 1, 462, 874, 515, 142, 19, 1, 1716, 3958, 2934, 1083, 218, 23, 1, 6435, 17548, 15694, 7266, 1955, 310, 27, 1, 24310, 76627, 80324, 44758, 15086, 3195, 418, 31, 1, 92378, 330818, 397923, 259356, 105102, 27866, 4867, 542, 35, 1
Offset: 1
Triangle begins as:
1;
3, 1;
10, 7, 1;
35, 38, 11, 1;
126, 187, 82, 15, 1;
462, 874, 515, 142, 19, 1;
1716, 3958, 2934, 1083, 218, 23, 1;
6435, 17548, 15694, 7266, 1955, 310, 27, 1;
24310, 76627, 80324, 44758, 15086, 3195, 418, 31, 1;
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A046658:= func< n,k | Binomial(n,k)*(Binomial(n+1,2)*Catalan(n )/Catalan(k-1) -4^(n-k+1)*Binomial(k,2))/(n*(n-k+1)) >;
[A046658(n,k): k in [1..n], n in [1..12]]; // G. C. Greubel, Jul 28 2024
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T[n_, k_]:= (1/2)*Binomial[n,k-1]*(Binomial[2*n,n]/Binomial[2*(k-1), k -1] - 4^(n-k+1)*(k-1)/n);
Table[T[n, k], {n,12}, {k,n}]//Flatten (* G. C. Greubel, Jul 28 2024 *)
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def A046658(n,k): return (1/2)*binomial(n,k-1)*(binomial(2*n, n)/binomial(2*(k-1), k-1) - 4^(n-k+1)*(k-1)/n)
flatten([[A046658(n,k) for k in range(1,n+1)] for n in range(1,13)]) # G. C. Greubel, Jul 28 2024
Showing 1-3 of 3 results.
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