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A249151
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Largest m such that m! divides the product of elements on row n of Pascal's triangle: a(n) = A055881(A001142(n)).
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20
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1, 1, 2, 1, 4, 2, 6, 1, 2, 4, 10, 7, 12, 6, 4, 1, 16, 2, 18, 4, 6, 10, 22, 11, 4, 12, 2, 6, 28, 25, 30, 1, 10, 16, 6, 36, 36, 18, 12, 40, 40, 6, 42, 10, 23, 22, 46, 19, 6, 4, 16, 12, 52, 2, 10, 35, 18, 28, 58, 47, 60, 30, 63, 1, 12, 10, 66, 16, 22, 49, 70, 41, 72, 36, 4, 18, 10, 12, 78, 80, 2
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OFFSET
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0,3
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COMMENTS
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A000225 gives the positions of ones.
A006093 seems to give all such k, that a(k) = k.
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LINKS
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FORMULA
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EXAMPLE
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Binomial coeff. Their product Largest k!
Row 0 1 1 1!
Row 1 1 1 1 1!
Row 2 1 2 1 2 2!
Row 3 1 3 3 1 9 1!
Row 4 1 4 6 4 1 96 4! (96 = 4*24)
Row 5 1 5 10 10 5 1 2500 2! (2500 = 1250*2)
Row 6 1 6 15 20 15 6 1 162000 6! (162000 = 225*720)
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PROG
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(PARI)
A249151(n) = { my(uplim, padicvals, b); uplim = (n+3); padicvals = vector(uplim); for(k=0, n, b = binomial(n, k); for(i=1, uplim, padicvals[i] += valuation(b, prime(i)))); k = 1; while(k>0, for(i=1, uplim, if((padicvals[i] -= valuation(k, prime(i))) < 0, return(k-1))); k++); };
\\ Alternative implementation:
A001142(n) = prod(k=1, n, k^((k+k)-1-n));
A055881(n) = { my(i); i=2; while((0 == (n%i)), n = n/i; i++); return(i-1); }
for(n=0, 4096, write("b249151.txt", n, " ", A249151(n)));
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CROSSREFS
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Cf. A249423 (numbers k such that a(k) = k+1).
Cf. A249429 (numbers k such that a(k) > k).
Cf. A249433 (numbers k such that a(k) < k).
Cf. A249434 (numbers k such that a(k) >= k).
Cf. A249424 (numbers k such that a(k) = (k-1)/2).
Cf. A249428 (and the corresponding values, i.e. numbers n such that A249151(2n+1) = n).
Cf. A001142, A006093, A000225, A007917, A055881, A187059, A249346, A249421, A249430, A249431, A249432.
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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