A249150 -id:A249150 - cp's OEIS frontend

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

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

Antti Karttunen, Oct 28 2014

nonn

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

Antti Karttunen, Table of n, a(n) for n = 1..732

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.

A249426 Record values in A249150.

Original entry on oeis.org

0, 1, 3, 5, 9, 11, 15, 17, 21, 27, 29, 35, 39, 41, 45, 51, 57, 59, 62, 65, 69, 71, 77, 79, 81, 83, 87, 89, 95, 99, 101, 104, 105, 107, 111, 118, 125, 129, 131, 135, 137, 143, 147, 149, 153, 155, 159, 161, 164, 165, 167, 171, 177, 179, 181, 189, 191, 194, 195, 197, 199, 207, 209, 219, 221, 225, 227, 231, 237, 239, 249, 255

Offset: 1

Antti Karttunen, Oct 28 2014

nonn

Antti Karttunen, Table of n, a(n) for n = 1..999

One less than A249427.
Cf. A249150, A249425.
Differs from A040976 a(n) = prime(n) - 2 for the first time at n=19, where a(n) = 62, while A040976(19) = 65. And larger terms differs only a few times.

(define (A249426 n) (A249150 (A249425 n)))

a(n) = A249150(A249425(n)).

a(n) = A249427(n) - 1.

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

Antti Karttunen, Oct 25 2014

A000225 gives the positions of ones.

A006093 seems to give all such k, that a(k) = k.

			              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)

Chai Wah Wu, Table of n, a(n) for n = 0..10000 (terms 0..4096 from Antti Karttunen)

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.

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)));

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

(define (A249151 n) (A055881 (A001142 n)))

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

A187059 The exponent of highest power of 2 dividing the product of the elements of the n-th row of Pascal's triangle (A001142).

Original entry on oeis.org

0, 0, 1, 0, 5, 2, 4, 0, 17, 10, 12, 4, 18, 8, 11, 0, 49, 34, 36, 20, 42, 24, 27, 8, 58, 36, 39, 16, 47, 22, 26, 0, 129, 98, 100, 68, 106, 72, 75, 40, 122, 84, 87, 48, 95, 54, 58, 16, 162, 116, 119, 72, 127, 78, 82, 32, 147, 94, 98, 44, 108, 52, 57, 0, 321, 258, 260, 196, 266, 200, 203, 136, 282, 212, 215, 144, 223, 150, 154, 80, 322, 244, 247, 168, 255, 174, 178, 96, 275, 190, 194, 108, 204, 116, 121, 32, 418, 324, 327, 232, 335

Offset: 0

Bruce Reznick, Mar 05 2011

The exponent of the highest power of 2 which divides Product_{k=0..n} binomial(n, k). This can be computed using de Polignac's formula.

This is the function ord_2(Ḡ_n) extensively studied in Lagarias-Mehta (2014), and plotted in Fig. 1.1. - Antti Karttunen, Oct 22 2014

			For example, if n = 4, the power of 2 that divides 1*4*6*4*1 is 5.

I. Niven, H. S. Zuckerman, H. L. Montgomery, An Introduction to the Theory of Numbers, Wiley, 1991, pages 182, 183, 187 (Ex. 34).

Antti Karttunen and Paul Tek, Table of n, a(n) for n = 0..8191 (First 4096 terms from Karttunen)
J. Lagarias, Products of binomial coefficients and Farey fractions, Lecture in DIMACS Conference on Challenges of Identifying Integer Sequences, October 9, 2014; Part 1, Part 2.
Jeffrey C. Lagarias and Harsh Mehta, Products of binomial coefficients and unreduced Farey fractions, arXiv:1409.4145 [math.NT], 2014.

Row sums of triangular table A065040.
Row 1 of array A249421.
Cf. A000295 (a(2^k-2)), A000337 (a(2^k)), A005803 (a(2^k-3)), A036799 (a(2^k+1)), A109363 (a(2^k-4)).
Cf. A000178, A000788, A001142, A002109, A007318, A007814, A174605, A249152, A249150, A249151, A249154, A249343, A249345, A249346, A249347.

a187059 = a007814 . a001142  -- Reinhard Zumkeller, Mar 16 2015

a[n_] := Sum[IntegerExponent[Binomial[n, k], 2], {k, 0, n}]; Array[a, 100, 0]
A187059[n_] := Sum[#*((#+1)*2^k - n - 1) & [Floor[n/2^k]], {k, Floor[Log2[n]]}];
Array[A187059, 100, 0] (* Paolo Xausa, Feb 11 2025 *)
2*Accumulate[#] - Range[Length[#]]*# & [DigitCount[Range[0, 99], 2, 1]] (* Paolo Xausa, Feb 11 2025 *)

a(n)=sum(k=0,n,valuation(binomial(n,k),2))

\\ Much faster version, based on code for A065040 by Charles R Greathouse IV which if reduced even further gives the formula a(n) = 2*A000788(n) - A249154(n):
A065040(m,k) = (hammingweight(k)+hammingweight(m-k)-hammingweight(m));
A187059(n) = sum(k=0, n, A065040(n, k));
for(n=0, 4095, write("b187059.txt", n, " ", A187059(n)));
\\ Antti Karttunen, Oct 25 2014

def A187059(n): return (n+1)*n.bit_count()+sum((m:=1<>j)-(r if n<<1>=m*(r:=k<<1|1) else 0)) for j in range(1,n.bit_length()+1)) # Chai Wah Wu, Nov 11 2024

a(2^k-1) = 0 (19th century); a(2^k) = (k-1)*2^k+1 for k >= 1. (Use de Polignac.)
a(n) = Sum_{i=0..n} A065040(n,i) [where the entries of triangular table A065040(m,k) give the exponent of the maximal power of 2 dividing binomial coefficient A007318(m,k)].
a(n) = A007814(A001142(n)). - Jason Kimberley, Nov 02 2011
a(n) = A249152(n) - A174605(n). [Exponent of 2 in the n-th hyperfactorial minus exponent of 2 in the n-th superfactorial. Cf. for example Lagarias & Mehta paper or Peter Luschny's formula for A001142.] - Antti Karttunen, Oct 25 2014
a(n) = 2*A000788(n) - A249154(n). - Antti Karttunen, Nov 02 2014
a(n) = Sum_{i=1..n} (2*i-n-1)*v_2(i), where v_2(i) = A007814(i) is the exponent of the highest power of 2 dividing i. - Ridouane Oudra, Jun 02 2022
a(n) = Sum_{k=1..floor(log_2(n))} t*((t+1)*2^k - n - 1), where t = floor(n/(2^k)). - Paolo Xausa, Feb 11 2025, derived from Ridouane Oudra's formula above.

Name clarified by Antti Karttunen, Oct 22 2014

A249070 a(n+1) gives the number of occurrences of the maximum digit of a(n) in factorial base (i.e., A246359(a(n))) so far amongst the factorial base representations of all the terms up to and including a(n), with a(0)=0.

Original entry on oeis.org

0, 1, 1, 2, 3, 5, 1, 7, 9, 12, 2, 13, 3, 16, 5, 6, 18, 1, 19, 2, 21, 3, 25, 27, 30, 32, 35, 7, 40, 9, 44, 4, 10, 11, 12, 13, 14, 15, 16, 18, 5, 19, 6, 56, 20, 7, 61, 22, 8, 64, 26, 66, 9, 69, 10, 29, 30, 76, 11, 32, 81, 12, 33, 88, 13, 36, 37, 38, 39, 40, 42, 14, 43, 15, 44, 16, 46, 17, 49, 50, 51, 52

Offset: 0

Antti Karttunen, Oct 20 2014

			   a(0) =  0 (by definition)
   a(1) =  1 ('1' in factorial base), as 0 has occurred once in all the preceding terms.
   a(2) =  1 as 1 has occurred once in all the preceding terms.
   a(3) =  2 ('10' in factorial base), as digit '1' has occurred two times in total in all the preceding terms.
   a(4) =  3 ('11' in factorial base), as '1' occurs once in each a(1) and a(2) and a(3).
   a(5) =  5 ('21' in factorial base), as '1' occurs once in each of a(1), a(2) and a(3) and twice at a(4).
   a(6) =  1 ('1' in factorial base), as '2' so far occurs only once at a(5)
   a(7) =  7 = '101'
   a(8) =  9 = '111'
   a(9) = 12 = '200'
  a(10) =  2 = '2'
  a(11) = 13 = '201'
  a(12) =  3 = '11'
  a(12) =  3 = '11'
  a(13) = 16 = '220'
  a(14) =  5 = '21'
  a(15) =  6 = '100'
  a(16) = 18 = '300'
  a(17) =  1 = '1'
  a(18) = 19 = '301'
  a(19) =  2 = '10'
  a(20) = 21 = '311'
  a(21) =  3 = '11'
  a(22) = 25 = '1001'
  a(23) = 27 = '1011'
  a(24) = 30 = '1100'
  a(25) = 32 = '1110'
  a(26) = 35 = '1121'
  a(27) =  7 (= '101' in factorial base), as the maximum digit in the factorial base representation of a(26), namely '2', has occurred in total 7 times in terms a(0) - a(26): once in each of a(5), a(9), a(11), a(14) and a(26), and twice in a(13).

Antti Karttunen, Table of n, a(n) for n = 0..10080

Cf. A007623, A246359, A249146, A249150.

Differs from A249069 for the first time at n=27, where a(27) = 7, while A249069(27) = 38.

cp's OEIS Frontend

A249425 Positions of records in A249150 and A249151.

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

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A249426 Record values in A249150.

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A249151 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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A187059 The exponent of highest power of 2 dividing the product of the elements of the n-th row of Pascal's triangle (A001142).

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A249070 a(n+1) gives the number of occurrences of the maximum digit of a(n) in factorial base (i.e., A246359(a(n))) so far amongst the factorial base representations of all the terms up to and including a(n), with a(0)=0.

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