cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A249151 Largest m such that m! divides the product of elements on row n of Pascal's triangle: a(n) = A055881(A001142(n)).

Original entry on oeis.org

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
Offset: 0

Views

Author

Antti Karttunen, Oct 25 2014

Keywords

Comments

A000225 gives the positions of ones.
A006093 seems to give all such k, that a(k) = k.

Examples

			              Binomial coeff.   Their product  Largest k!
                 A007318          A001142(n)   which divides
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)

Links

Crossrefs

One more than A249150.
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. A249425 (record positions).
Cf. A249427 (record values).
Cf. A001142, A006093, A000225, A007917, A055881, A187059, A249346, A249421, A249430, A249431, A249432.

Programs

  • 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); }
A249151(n) = A055881(A001142(n));
for(n=0, 4096, write("b249151.txt", n, " ", A249151(n)));
  • Python
    from itertools import count
from collections import Counter
from math import comb
from sympy import factorint
def A249151(n):
    p = sum((Counter(factorint(comb(n,i))) for i in range(n+1)),start=Counter())
    for m in count(1):
        f = Counter(factorint(m))
        if not f<=p:
            return m-1
        p -= f # Chai Wah Wu, Aug 19 2025
  • Scheme
    (define (A249151 n) (A055881 (A001142 n)))

Formula

a(n) = A055881(A001142(n)).

A249434 Integers m such that m! divides the product of elements on row m of Pascal's triangle.

Original entry on oeis.org

0, 1, 2, 4, 6, 10, 12, 16, 18, 22, 28, 30, 35, 36, 39, 40, 42, 46, 52, 58, 60, 62, 66, 70, 72, 78, 79, 82, 83, 88, 89, 96, 100, 102, 104, 106, 107, 108, 112, 126, 130, 131, 136, 138, 143, 148, 149, 150, 153, 156, 159, 162, 164, 166, 167, 172, 174, 175, 178, 179, 180, 181, 190, 192, 194, 196, 197, 198, 199, 207, 209, 210, 219, 222, 226, 228, 232, 238, 240, 250, 256
Offset: 1

Views

Author

Antti Karttunen, Nov 02 2014

Keywords

Comments

Integers m such that A249151(m) >= m.
Equally: Integers m such that A249431(m) is nonnegative.
It seems that A006093 gives all those k for which A249151(k) = k. If that is true, then this is a disjoint union of A006093 and A249429.

Examples

			0! = 1 divides the product of binomial coefficients on row 0 of A007318, namely {1}, thus a(1) = 0.
1! = 1 divides the product of row 1 (1*1), thus a(2) = 1.
2! = 2 divides the product of row 2 (1*2*1), thus a(3) = 2.
3! = 6 does not divide the product of row 3 (1*3*3*1), but 4! = 24 divides the product of row 4 (1*4*6*4*1), as 96 = 4*24, thus a(4) = 4.

Links

Crossrefs

Complement: A249433.
Subsequences: A006093 (conjectured), A249429, A249430, A249432.
Cf. A000142, A001142, A007318, A249151, A249431.

A249423 Integers n such that A249150(n) = n; integers n such that A249151(n) = n+1.

Original entry on oeis.org

0, 35, 39, 62, 79, 83, 89, 104, 107, 131, 143, 149, 153, 159, 164, 167, 175, 179, 181, 194, 197, 199, 207, 209, 219, 259, 263, 269, 272, 274, 279, 285, 287, 296, 299, 305, 307, 311, 314, 319, 329, 339, 356, 359, 363, 373, 377, 379, 384, 389, 391, 395, 399, 407, 415, 417, 419, 424, 428, 431, 441, 449, 455, 461, 467, 475, 489, 512
Offset: 1

Views

Author

Antti Karttunen, Oct 28 2014

Keywords

Comments

Integers n such that {product of elements on row n of Pascal's triangle} is divisible by (n+1)! but not by (n+2)!

Links

Crossrefs

Subsequence of A249434 and of A249429; it differs from the latter for the first time at n=17, where a(17) = 175 > 174 = A249429(17).
Cf. A000142, A001142, A007318, A249150, A249151, A249424, A249425, A249426.
Showing 1-3 of 3 results.

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