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A075833
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a(n) is the least k such that for any p prime dividing n, p does not divide binomial((n+1)*k, k+1), or -1 if no such k exists.
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0
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1, 1, 2, 3, 4, 11, 6, 7, 8, 29, 10, 19847, 12, 55, 29, 15, 16, 266831, 18, 259, 62, 131, 22, 71, 24, 519, 26, 55, 28
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OFFSET
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1,3
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COMMENTS
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The initial terms (with a question mark for currently unknown terms) are 1, 1, 2, 3, 4, 11, 6, 7, 8, 29, 10, 19847, 12, 55, 29, 15, 16, 266831, 18, 259, 62, 131, 22, 71, 24, 519, 26, 55, 28, ?, 30, 31, 32, 305, 34, 536579, 36, 2203, 545219, 39, 40, 140069, 42, 2067, 89, 3219, 46, 4655, 48, 328799, 305, 207, 52, 70739, 274, 356383, 398, 6785, 58, ?, ... .
It seems that a(30) = -1, otherwise a(30) > 10^10. To prove it one needs to show that for every k, binomial(31*k,k+1) is divisible by 2, 3, or 5. The next values of n for which a(n) = -1 seem to be 60, 66, 78, 84, 90, 105, ... . - Pontus von Brömssen, Jan 30 2022
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LINKS
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FORMULA
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a(p^m) = p^m - 1 for prime p and m > 0. [Proof: if k in base p is x_1 x_2 ... x_t and t <= m, then (p^m+1)*k in base p is x_1 x_2 ... x_t 0 0 ... 0 x_1 x_2 ... x_t. Let k+1 in base p be y_1 y_2 ... y_r, where r = t or t+1. By Lucas's theorem, we have y_r <= x_t, y_(r-1) <= x_(t-1), y_(r-2) <= x_(t-2), ... Therefore, x_1 = x_2 = ... = x_m = p-1 and k in base 10 is p^m - 1. - Jinyuan Wang, Apr 06 2020]
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PROG
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(PARI) D(k, n) = binomial((n+1)*k, k+1);
a(n) = {my(d=divisors(n), k=1); while(prod(i=1, numdiv(n), D(k, n)%if(isprime(component(d, i)), component(d, i), D(k, n)+1)) == 0, k++); k; }
(PARI) isok(x, f) = for (i=1, #f, if (!(x % f[i]), return(0))); return(1);
a(n) = my(k=1, f=factor(n)[, 1]~); while (!isok(binomial((n+1)*k, k+1), f), k++); k; \\ Michel Marcus, Jan 29 2022
(PARI) f(x, k) = if (x, x\k + f(x\k, k)); \\ valuation(x!, k)
isoki(x, y, k) = f(x, k) - f(y, k) - f(x-y, k) == 0;
isokf(x, y, f) = for (i=1, #f, if (! isoki(x, y, f[i]), return(0))); return(1);
af(n) = my(k=1, f=factor(n)[, 1]~); while (!isokf((n+1)*k, k+1, f), k++); k; \\ Michel Marcus, Jan 30 2022
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CROSSREFS
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KEYWORD
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nonn,more
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AUTHOR
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EXTENSIONS
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Edited by N. J. A. Sloane, Jan 29 2022: the previous definition was unacceptable; changed escape clause value to -1; deleted terms starting at a(18).
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STATUS
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approved
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