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%I #57 Feb 02 2022 21:26:04
%S 1,1,2,3,4,11,6,7,8,29,10,19847,12,55,29,15,16,266831,18,259,62,131,
%T 22,71,24,519,26,55,28
%N 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.
%C 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, ?, ... .
%C a(36) = 536579. - _Michel Marcus_, Jan 30 2022
%C 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
%C Either a(30) = -1 or a(30) > 10^17. - _Charles R Greathouse IV_, Feb 02 2022
%F 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]
%o (PARI) D(k,n) = binomial((n+1)*k,k+1);
%o 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; }
%o (PARI) isok(x, f) = for (i=1, #f, if (!(x % f[i]), return(0))); return(1);
%o 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
%o (PARI) f(x, k) = if (x, x\k + f(x\k, k)); \\ valuation(x!, k)
%o isoki(x, y, k) = f(x, k) - f(y, k) - f(x-y, k) == 0;
%o isokf(x, y, f) = for (i=1, #f, if (! isoki(x,y,f[i]), return(0))); return(1);
%o af(n) = my(k=1, f=factor(n)[,1]~); while (!isokf((n+1)*k, k+1, f), k++); k; \\ _Michel Marcus_, Jan 30 2022
%K nonn,more
%O 1,3
%A _Benoit Cloitre_, Oct 14 2002
%E a(12), a(38) corrected by _Michel Marcus_, Jan 28 2022
%E 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).
%E a(18) from _Michel Marcus_, Jan 30 2022
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