A103214 -id:A103214 - cp's OEIS frontend

A287326 Triangle read by rows: T(n, k) = 6k(n-k) + 1; n >= 0, 0 <= k <= n.

1, 1, 1, 1, 7, 1, 1, 13, 13, 1, 1, 19, 25, 19, 1, 1, 25, 37, 37, 25, 1, 1, 31, 49, 55, 49, 31, 1, 1, 37, 61, 73, 73, 61, 37, 1, 1, 43, 73, 91, 97, 91, 73, 43, 1, 1, 49, 85, 109, 121, 121, 109, 85, 49, 1, 1, 55, 97, 127, 145, 151, 145, 127, 97, 55, 1, 1, 61, 109, 145, 169, 181, 181, 169, 145, 109, 61, 1

Offset: 0

Kolosov Petro, Aug 31 2017

From Kolosov Petro, Apr 12 2020: (Start)
Let A(m, r) = A302971(m, r) / A304042(m, r).
Let L(m, n, k) = Sum_{r=0..m} A(m, r) * k^r * (n - k)^r.
Then T(n, k) = L(1, n, k), i.e T(n, k) is partial case of L(m, n, k) for m = 1.
T(n, k) is symmetric: T(n, k) = T(n, n-k). (End)

			Triangle begins:
  ----------------------------------------
  k=    0   1   2   3   4   5   6   7   8
  ----------------------------------------
  n=0:  1;
  n=1:  1,  1;
  n=2:  1,  7,  1;
  n=3:  1, 13, 13,  1;
  n=4:  1, 19, 25, 19,  1;
  n=5:  1, 25, 37, 37, 25,  1;
  n=6:  1, 31, 49, 55, 49, 31,  1;
  n=7:  1, 37, 61, 73, 73, 61, 37,  1;
  n=8:  1, 43, 73, 91, 97, 91, 73, 43,  1;

Georg Fischer, Table of n, a(n) for n = 0..495 [rows 0..10 and 12..30 from Kolosov Petro]
Petro Kolosov, On the link between binomial theorem and discrete convolution, arXiv:1603.02468 [math.NT], 2016-2025.
Petro Kolosov, Polynomial identity involving binomial theorem and Faulhaber's formula, 2023.
Petro Kolosov, History and overview of the polynomial P_b^m(x), 2024.

Columns k=0..6 give A000012, A016921, A017533, A161705, A103214, A128470, A158065.
Column sums k=0..4 give A000027, A000567, A051866, A051872, A255185.
Row sums give A001093.
Various cases of L(m, n, k): This sequence (m=1), A300656(m=2), A300785(m=3). See comments for L(m, n, k).
Differences of cubes n^3 are T(A000124(n), 1).
Cf. A000124, A000578, A003154, A003215, A007318, A008458, A016969.
Cf. A038593, A055012, A068601, A077028, A094053, A166873, A227776.
Cf. A294317, A302971, A304042.

Flat(List([0..11],n->List([0..n],k->6*k*(n-k)+1))); # Muniru A Asiru, Oct 09 2018

/* As triangle */ [[6*k*(n-k) + 1: k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Oct 26 2018

T := (n, k) -> 6*k*(n-k) + 1:
seq(seq(T(n, k), k=0..n), n=0..11); # Muniru A Asiru, Oct 09 2018

T[n_, k_] := 6 k (n - k) + 1; Column[Table[T[n, k], {n, 0, 10}, {k, 0, n}], Center] (* Kolosov Petro, Jun 02 2019 *)

t(n, k) = 6*k*(n-k)+1
trianglerows(n) = for(x=0, n-1, for(y=0, x, print1(t(x, y), ", ")); print(""))
/* Print initial 9 rows of triangle as follows */
trianglerows(9) \\ Felix Fröhlich, Jan 09 2018

def A287326(n,k): return 6*k*(n-k) + 1
flatten([[A287326(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Sep 25 2024

T(n, k) = 6*k*(n-k) + 1.
G.f. of column k: n^k*(1+(6*k-1)*n)/(1-n)^2.
G.f.: (1 - x - x*y + 7*x^2*y)/((1 - x)^2*(1 - x*y)^2). - Stefano Spezia, Oct 09 2018 [Adapted by Stefano Spezia, Sep 25 2024]
From Kolosov Petro, Jun 05 2019: (Start)
T(n, k) = 1/2 * T(A294317(n, k), k) + 1/2.
T(n+1, k) = 2*T(n, k) - T(n-1, k), for n >= k.
T(n, k) = 6*A077028(n, k) - 5.
T(2n, n) = A227776(n).
T(2n+1, n) = A003154(n+1).
T(2n+3, n) = A166873(n+1).
Sum_{k=0..n-1} T(n, k) = Sum_{k=1..n} T(n, k) = A000578(n).
Sum_{k=1..n-1} T(n, k) = A068601(n).
(n+1)^3 - n^3 = T(A000124(n), 1). (End)
Sum_{k=0..n} (-1)^k*T(n, k) = (-1/2)*(1 + (-1)^n)*A016969(floor(n/2) - 1). - G. C. Greubel, Sep 25 2024

A334832 Numbers whose squarefree part is congruent to 1 (mod 24).

Original entry on oeis.org

1, 4, 9, 16, 25, 36, 49, 64, 73, 81, 97, 100, 121, 144, 145, 169, 193, 196, 217, 225, 241, 256, 265, 289, 292, 313, 324, 337, 361, 385, 388, 400, 409, 433, 441, 457, 481, 484, 505, 529, 553, 576, 577, 580, 601, 625, 649, 657, 673, 676, 697, 721, 729, 745, 769, 772, 784, 793, 817, 841

Offset: 1

Peter Munn, Jun 15 2020

Closed under multiplication.
The sequence forms a subgroup of the positive integers under the commutative operation A059897(.,.). A059897 has a relevance to squarefree parts that arises from the identity A007913(k*m) = A059897(A007913(k), A007913(m)), where A007913(n) is the squarefree part of n.
The subgroup is one of 8 A059897(.,.) subgroups of the form {k : A007913(k) == 1 (mod m)}. The list seems complete, in anticipation of proof that such sets form subgroups only when m is a divisor of 24 (based on the property described by A. G. Astudillo in A018253). This sequence might be viewed as primitive with respect to the other 7, as the latter correspond to subgroups of its quotient group, in the sense that each one (as a set) is also a union of cosets described below. The 7 include A003159 (m=2), A055047 (m=3), A277549 (m=4), A234000 (m=8) and the trivial A000027 (m=1).
The subgroup has 32 cosets. For each i in {1, 5, 7, 11, 13, 17, 19, 23}, j in {1, 2, 3, 6} there is a coset {n : n = k^2 * (24m + i) * j, k >= 1, m >= 0}. The divisors of 2730 = 2*3*5*7*13 form a transversal. (11, clearly not such a divisor, is in the same coset as 35 = 11 + 24; 17, 19, 23 are in the same cosets as 65, 91, 455 respectively.)
The asymptotic density of this sequence is 1/16. - Amiram Eldar, Mar 08 2021

			The squarefree part of 26 is 26, which is congruent to 2 (mod 24), so 26 is not in the sequence.
The squarefree part of 292 = 2^2 * 73 is 73, which is congruent to 1 (mod 24), so 292 is in the sequence.

Amiram Eldar, Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, Quotient Group.
Eric Weisstein's World of Mathematics, Right Transversal.

A subgroup under A059897, defined using A007913.
Subsequences: A000290\{0}, A103214, A107008.
Equivalent sequences modulo other members of A018253: A000027 (1), A003159 (2), A055047 (3), A277549 (4), A352272(6), A234000 (8).

Select[Range[850], Mod[Sqrt[#] /. (c_ : 1)*a_^(b_ : 0) :> (c*a^b)^2, 24] == 1 &] (* Michael De Vlieger, Jun 24 2020 *)

isok(m) = core(m) % 24 == 1; \\ Peter Munn, Jun 21 2020

from sympy import integer_log
def A334832(n):
    def bisection(f,kmin=0,kmax=1):
        while f(kmax) > kmax: kmax <<= 1
        kmin = kmax >> 1
        while kmax-kmin > 1:
            kmid = kmax+kmin>>1
            if f(kmid) <= kmid:
                kmax = kmid
            else:
                kmin = kmid
        return kmax
    def f(x): return n+x-sum((x//(i<<(j<<1))-1)//24+1 for i in (9**k for k in range(integer_log(x,9)[0]+1)) for j in range((x//i>>1).bit_length()+1))
    return bisection(f,n,n) # Chai Wah Wu, Mar 21 2025

{a(n)} = {n : n = k^2 * (24m + 1), k >= 1, m >= 0}.

A215148 a(n) = 23*n + 1.

Original entry on oeis.org

1, 24, 47, 70, 93, 116, 139, 162, 185, 208, 231, 254, 277, 300, 323, 346, 369, 392, 415, 438, 461, 484, 507, 530, 553, 576, 599, 622, 645, 668, 691, 714, 737, 760, 783, 806, 829, 852, 875, 898, 921, 944, 967, 990, 1013, 1036, 1059, 1082, 1105, 1128, 1151, 1174

Offset: 0

Jeremy Gardiner, Aug 04 2012

Jeremy Gardiner, Table of n, a(n) for n = 0..999
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A103214, A120344, A161709.

I:=[1,24]; [n le 2 select I[n] else 2*Self(n-1) - Self(n-2): n in [1..30]]; // G. C. Greubel, Apr 19 2018

Range[1, 1000, 23]
LinearRecurrence[{2,-1}, {1,24}, 50] (* G. C. Greubel, Apr 19 2018 *)

for(n=0, 50, print1(23*n + 1, ", ")) \\ G. C. Greubel, Apr 19 2018

From G. C. Greubel, Apr 19 2018: (Start)
a(n) = 2*a(n-1) - a(n-2).
G.f.: (1+22*x)/(1-x)^2.
E.g.f.: (23*x + 1)*exp(x). (End)

A350052 Third part of the trisection of A017077: a(n) = 17 + 24*n.

Original entry on oeis.org

17, 41, 65, 89, 113, 137, 161, 185, 209, 233, 257, 281, 305, 329, 353, 377, 401, 425, 449, 473, 497, 521, 545, 569, 593, 617, 641, 665, 689, 713, 737, 761, 785, 809, 833, 857, 881, 905, 929, 953, 977, 1001, 1025, 1049, 1073

Offset: 0

Wolfdieter Lang, Dec 11 2021

The trisection of A017077 = {1 + 8*k}A103214%20=%20%7B1%20+%2024*n%7D">{k>=0} gives A103214 = {1 + 24*n}{n>=0}, 3*A017101 = {3*(3 + 8*n)}{n >= 0} and {a(n)}{n>=0}. These three sequences are congruent to 1 modulo 8 and to 1, 3, and 5 modulo 6, respectively.

Winston de Greef, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A008606, A017077, 3*A017101, A103214.

24 * Range[0, 44] + 17 (* Amiram Eldar, Dec 18 2021 *)

a(n) = 17 + 24*n \\ Winston de Greef, Jan 28 2024

a(n) = 17 + 24*n = 17 + A008606(n), for n >= 0
a(n) = 2*a(n-1) - a(n-2), for n >= 1, with a(-1) = -7, a(0) = 17.
G.f.: (17 + 7*x)/(1-x)^2.
E.g.f.: (17 + 24*x)*exp(x).

A214492 Arithmetic mean of next a(n) successive squares of positive integers is a square.

Original entry on oeis.org

337, 649, 961, 1273, 1585, 1897, 1919, 1151, 1223, 1295, 1367, 1439, 1511, 1583, 1655, 1727, 1799, 1871, 1943, 2015, 2087, 2159, 2231, 2303, 2375, 2447, 2519, 1487, 1511, 1535, 1559, 1583, 1607, 1631, 1655, 1679, 1703, 1727, 1751, 1775, 1799, 1823, 1847, 1871, 1895, 1919

Offset: 1

Alex Ratushnyak, Jul 19 2012

nonn

Only squares of positive integers, starting from 1; zero is not included. (If it were included, A103214 would result.)
Also, a(n)=1 is obviously not permitted: must be 2 or more successive squares, otherwise all a(n)=1.
Among first 1363 terms all are odd, 933 are primes, a(n) < a(n-1) twice.
Corresponding arithmetic means that are perfect squares:
b(n) = 38025, 473344, 2229049, 6812100, 16313521, 33408400, 59013124, 84695209, 107952100, 135699201, 168480400, 206870689, 251476164, 302934025, 361912576, 429111225, 505260484, 591121969, 687488400
Their square roots c(n) = sqrt(b(n)):
195, 688, 1493, 2610, 4039, 5780, 7682, 9203, 10390, 11649, 12980, 14383, 15858, 17405, 19024, 20715, 22478, 24313, 26220, 28199, 30250, 32373, 34568, 36835, 39174, 41585, 44068, 46067, 47566, 49089, 50636, 52207

			(1 + 4 + 9 + ... + 337^2)/337 = 38025, which is a square, so 337 is a term.
(338^2 + ... + (338 + 648)^2)/649 = 473344, which is a square, so 649 is a term.

Cf. A000290, A073684.

Cf. A103214, excluding first term: arithmetic mean of next a(n) successive squares of nonnegative integers is a square.

import math
sum = k = 0
for n in range(1, 220000):
    sum += n*n
    k += 1
    sqr = int(math.sqrt(sum*1.0/k))
    while sqr*sqr*ksum:
        sqr-=1
    if sqr*sqr*k==sum and k>1:
        print(k, end=',')
        sum = k = 0

cp's OEIS Frontend

A287326 Triangle read by rows: T(n, k) = 6*k*(n-k) + 1; n >= 0, 0 <= k <= n.

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A334832 Numbers whose squarefree part is congruent to 1 (mod 24).

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A215148 a(n) = 23*n + 1.

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A350052 Third part of the trisection of A017077: a(n) = 17 + 24*n.

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A214492 Arithmetic mean of next a(n) successive squares of positive integers is a square.

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A287326 Triangle read by rows: T(n, k) = 6k(n-k) + 1; n >= 0, 0 <= k <= n.