A209260 -id:A209260 - cp's OEIS frontend

A141419 Triangle read by rows: T(n, k) = A000217(n) - A000217(n - k) with 1 <= k <= n.

1, 2, 3, 3, 5, 6, 4, 7, 9, 10, 5, 9, 12, 14, 15, 6, 11, 15, 18, 20, 21, 7, 13, 18, 22, 25, 27, 28, 8, 15, 21, 26, 30, 33, 35, 36, 9, 17, 24, 30, 35, 39, 42, 44, 45, 10, 19, 27, 34, 40, 45, 49, 52, 54, 55

Offset: 1

Roger L. Bagula, Aug 05 2008

As a rectangle, the accumulation array of A051340.
From Clark Kimberling, Feb 05 2011: (Start)
Here all the weights are divided by two where they aren't in Cahn.
As a rectangle, A141419 is in the accumulation chain
... < A051340 < A141419 < A185874 < A185875 < A185876 < ...
(See A144112 for the definition of accumulation array.)
row 1: A000027
col 1: A000217
diag (1,5,...): A000326 (pentagonal numbers)
diag (2,7,...): A005449 (second pentagonal numbers)
diag (3,9,...): A045943 (triangular matchstick numbers)
diag (4,11,...): A115067
diag (5,13,...): A140090
diag (6,15,...): A140091
diag (7,17,...): A059845
diag (8,19,...): A140672
(End)
Let N=2*n+1 and k=1,2,...,n. Let A_{N,n-1} = [0,...,0,1; 0,...,0,1,1; ...; 0,1,...,1; 1,...,1], an n X n unit-primitive matrix (see [Jeffery]). Let M_n=[A_{N,n-1}]^4. Then t(n,k)=[M_n](1,k), that is, the n-th row of the triangle is given by the first row of M_n. - _L. Edson Jeffery, Nov 20 2011
Conjecture. Let N=2*n+1 and k=1,...,n. Let A_{N,0}, A_{N,1}, ..., A_{N,n-1} be the n X n unit-primitive matrices (again see [Jeffery]) associated with N, and define the Chebyshev polynomials of the second kind by the recurrence U_0(x) = 1, U_1(x) = 2*x and U_r(x) = 2*x*U_(r-1)(x) - U_(r-2)(x)  (r>1). Define the column vectors V_(k-1) = (U_(k-1)(cos(Pi/N)), U_(k-1)(cos(3*Pi/N)), ..., U_(k-1)(cos((2*n-1)*Pi/N)))^T, where T denotes matrix transpose. Let S_N = [V_0, V_1, ..., V_(n-1)] be the n X n matrix formed by taking V_(k-1) as column k-1. Let X_N = [S_N]^T*S_N, and let [X_N](i,j) denote the entry in row i and column j of X_N, i,j in {0,...,n-1}. Then t(n,k) = [X_N](k-1,k-1), and row n of the triangle is given by the main diagonal entries of X_N. Remarks: Hence t(n,k) is the sum of squares t(n,k) = sum[m=1,...,n (U_(k-1)(cos((2*m-1)*Pi/N)))^2]. Finally, this sequence is related to A057059, since X_N = [sum_{m=1,...,n} A057059(n,m)*A_{N,m-1}] is also an integral linear combination of unit-primitive matrices from the N-th set. - L. Edson Jeffery, Jan 20 2012
Row sums: n*(n+1)*(2*n+1)/6. - L. Edson Jeffery, Jan 25 2013
n-th row = partial sums of n-th row of A004736. - Reinhard Zumkeller, Aug 04 2014
T(n,k) is the number of distinct sums made by at most k elements in {1, 2, ... n}, for 1 <= k <= n, e.g., T(6,2) = the number of distinct sums made by at most 2 elements in {1,2,3,4,5,6}. The sums range from 1, to 5+6=11. So there are 11 distinct sums. - Derek Orr, Nov 26 2014
A number n occurs in this sequence A001227(n) times, the number of odd divisors of n, see A209260. - Hartmut F. W. Hoft, Apr 14 2016
Conjecture: 2*n + 1 is composite if and only if gcd(t(n,m),m) != 1, for some m. - L. Edson Jeffery, Jan 30 2018
From Peter Munn, Aug 21 2019 in respect of the sequence read as a triangle: (Start)
A number m can be found in column k if and only if A286013(m, k) is nonzero, in which case m occurs in column k on row A286013(m, k).
The first occurrence of m is in row A212652(m) column A109814(m), which is the rightmost column in which m occurs. This occurrence determines where m appears in A209260. The last occurrence of m is in row m column 1.
Viewed as a sequence of rows, consider the subsequences (of rows) that contain every positive integer. The lexicographically latest of these subsequences consists of the rows with row numbers in A270877; this is the only one that contains its own row numbers only once.
(End)

			As a triangle:
   1,
   2,  3,
   3,  5,  6,
   4,  7,  9, 10,
   5,  9, 12, 14, 15,
   6, 11, 15, 18, 20, 21,
   7, 13, 18, 22, 25, 27, 28,
   8, 15, 21, 26, 30, 33, 35, 36,
   9, 17, 24, 30, 35, 39, 42, 44, 45,
  10, 19, 27, 34, 40, 45, 49, 52, 54, 55;
As a rectangle:
   1   2   3   4   5   6   7   8   9  10
   3   5   7   9  11  13  15  17  19  21
   6   9  12  15  18  21  24  27  30  33
  10  14  18  22  26  30  34  38  42  46
  15  20  25  30  35  40  45  50  55  60
  21  27  33  39  45  51  57  63  69  75
  28  35  42  49  56  63  70  77  84  91
  36  44  52  60  68  76  84  92 100 108
  45  54  63  72  81  90  99 108 117 126
  55  65  75  85  95 105 115 125 135 145
Since the odd divisors of 15 are 1, 3, 5 and 15, number 15 appears four times in the triangle at t(3+(5-1)/2, 5) in column 5 since 5+1 <= 2*3, t(5+(3-1)/2, 3), t(1+(15-1)/2, 2*1) in column 2 since 15+1 > 2*1, and t(15+(1-1)/2, 1). - _Hartmut F. W. Hoft_, Apr 14 2016

R. N. Cahn, Semi-Simple Lie Algebras and Their Representations, Dover, NY, 2006, ISBN 0-486-44999-8, p. 139.

Reinhard Zumkeller, Rows n = 1..125 of triangle, flattened
Johann Cigler, Some elementary observations on Narayana polynomials and related topics, arXiv:1611.05252 [math.CO], 2016. See p. 24.
Carlton Gamer, David W. Roeder, and John J. Watkins, Trapezoidal Numbers, Mathematics Magazine 58:2 (1985), pp. 108-110.
L. E. Jeffery, Unit-primitive matrices
M. A. Nyblom, On the representation of the integers as a difference of nonconsecutive triangular numbers, Fibonacci Quarterly 39:3 (2001), pp. 256-263.

Cf. A000330 (row sums), A004736, A057059, A070543.
A144112, A051340, A141419, A185874, A185875, A185876 are accumulation chain related.
A141418 is a variant.
Cf. A001227, A209260. - Hartmut F. W. Hoft, Apr 14 2016
A109814, A212652, A270877, A286013 relate to where each natural number appears in this sequence.
A000027, A000217, A000326, A005449, A045943, A059845, A115067, A140090, A140091, A140672 are rows, columns or diagonals - refer to comments.

a141419 n k =  k * (2 * n - k + 1) `div` 2
a141419_row n = a141419_tabl !! (n-1)
a141419_tabl = map (scanl1 (+)) a004736_tabl
-- Reinhard Zumkeller, Aug 04 2014

a:=(n,k)->k*n-binomial(k,2): seq(seq(a(n,k),k=1..n),n=1..12); # Muniru A Asiru, Oct 14 2018

T[n_, m_] = m*(2*n - m + 1)/2; a = Table[Table[T[n, m], {m, 1, n}], {n, 1, 10}]; Flatten[a]

t(n,m) = m*(2*n - m + 1)/2.
t(n,m) = A000217(n) - A000217(n-m). - L. Edson Jeffery, Jan 16 2013
Let v = d*h with h odd be an integer factorization, then v = t(d+(h-1)/2, h) if h+1 <= 2*d, and v = t(d+(h-1)/2, 2*d) if h+1 > 2*d; see A209260. - Hartmut F. W. Hoft, Apr 14 2016
G.f.: y*(-x + y)/((-1 + x)^2*(-1 + y)^3). - Stefano Spezia, Oct 14 2018
T(n, 2) = A060747(n) for n > 1. T(n, 3) = A008585(n - 1) for n > 2. T(n, 4) = A016825(n - 2) for n > 3. T(n, 5) = A008587(n - 2) for n > 4. T(n, 6) = A016945(n - 3) for n > 5. T(n, 7) = A008589(n - 3) for n > 6. T(n, 8) = A017113(n - 4) for n > 7.r n > 5. T(n, 7) = A008589(n - 3) for n > 6. T(n, 8) = A017113(n - 4) for n > 7. T(n, 9) = A008591(n - 4) for n > 8. T(n, 10) = A017329(n - 5) for n > 9. T(n, 11) = A008593(n - 5) for n > 10.  T(n, 12) = A017593(n - 6) for n > 11. T(n, 13) = A008595(n - 6) for n > 12. T(n, 14) = A147587(n - 7) for n > 13. T(n, 15) = A008597(n - 7) for n > 14. T(n, 16) = A051062(n - 8) for n > 15. T(n, 17) = A008599(n - 8) for n > 16. - Stefano Spezia, Oct 14 2018
T(2*n-k, k) = A070543(n, k). - Peter Munn, Aug 21 2019

Simpler name by Stefano Spezia, Oct 14 2018

A212652 a(n) is the least positive integer M such that n = T(M) - T(k), for k an integer, 0 <= k <= M, where T(r) = A000217(r) is the r-th triangular number.

Original entry on oeis.org

1, 2, 2, 4, 3, 3, 4, 8, 4, 4, 6, 5, 7, 5, 5, 16, 9, 6, 10, 6, 6, 7, 12, 9, 7, 8, 7, 7, 15, 8, 16, 32, 8, 10, 8, 8, 19, 11, 9, 10, 21, 9, 22, 9, 9, 13, 24, 17, 10, 12, 11, 10, 27, 10, 10, 11, 12, 16, 30, 11, 31, 17, 11, 64, 11, 11, 34

Offset: 1

L. Edson Jeffery, Feb 14 2013

nonn

n = A000217(a(n)) - A000217(a(n) - A109814(n)).
Conjecture: n appears in row a(n) of A209260.
From Daniel Forgues, Jan 06 2016: (Start)
n = Sum_{i=k+1..M} i = T(M) - T(k) = (M-k)*(M+k+1)/2.
n = 2^m, m >= 0, iff M = n = 2^m and k = n - 1 = 2^m - 1. (Points on line with slope 1.) (Powers of 2 can't be the sum of consecutive numbers.)
n is odd prime iff k = M-2. Thus M = (n+1)/2 when n is odd prime. (Points on line with slope 1/2.) (Odd primes can't be the sum of more than 2 consecutive numbers.) (End)
If n = 2^m*p where p is an odd prime, then a(n) = 2^m + (p-1)/2. - Robert Israel, Jan 14 2016
This also expresses the following geometry: along a circle having (n) points on its circumference, a(n) expresses the minimum number of hops from a start point, in a given direction (CW or CCW), when each hop is increased by one, before returning to a visited point. For example, on a clock (n=12), starting at 12 (same as zero), the hops would lead to the points 1, 3, 6, 10 and then 3, which was already visited: 5 hops altogether, so a(12) = 5. - Joseph Rozhenko, Dec 25 2023
Conjecture: a(n) is the smallest of the largest parts of the partitions of n into consecutive parts. - Omar E. Pol, Jan 07 2025

			For n = 63, we have D(63) = {1,3,7,9,21,63}, B_63 = {11,12,13,22,32,63} and a(63) = min(11,12,13,22,32,63) = 11. Since A109814(63) = 9, T(11) - T(11-9) = T(11) - T(2) = 66 - 3 = 63.

David W. Wilson, Table of n, a(n) for n = 1..10000
Max Alekseyev, is this sequence interesting?, Sequence Fans Mailing List, Mar 31 2008.

Cf. A000217, A109814, A118235, A138796, A141419, A209260, A218621.

f:= n ->  min(map(t -> n/t + (t-1)/2,
numtheory:-divisors(n/2^padic:-ordp(n,2)))):
map(f, [$1..100]); # Robert Israel, Jan 14 2016

Table[Min[n/# + (# - 1)/2 &@ Select[Divisors@ n, OddQ]], {n, 67}] (* Michael De Vlieger, Dec 11 2015 *)

{ A212652(n) = my(m); m=2*n+1; fordiv(n/2^valuation(n,2), d, m=min(m,d+(2*n)\d)); (m-1)\2; } \\ Max Alekseyev, Mar 31 2008

a(n) = Min_{odd d|n} (n/d + (d-1)/2).
a(n) = A218621(n) + (n/A218621(n) - 1)/2.
a(n) = A109814(n) + A118235(n) - 1.

Reference to Max Alekseyev's 2008 proposal of this sequence added by N. J. A. Sloane, Nov 01 2014

A218621 a(n) = unique divisor d of n such that d + (n/d - 1)/2 is minimal and integral.

Original entry on oeis.org

1, 2, 1, 4, 1, 2, 1, 8, 3, 2, 1, 4, 1, 2, 3, 16, 1, 2, 1, 4, 3, 2, 1, 8, 5, 2, 3, 4, 1, 6, 1, 32, 3, 2, 5, 4, 1, 2, 3, 8, 1, 6, 1, 4, 5, 2, 1, 16, 7, 10, 3, 4, 1, 6, 5, 8, 3, 2, 1, 4, 1, 2, 7, 64, 5, 6, 1, 4, 3, 10, 1, 8, 1, 2, 5, 4, 7, 6, 1, 16, 9, 2, 1, 4, 5

Offset: 1

L. Edson Jeffery, Feb 18 2013

nonn

Differs from A079891 starting at a(18).
For integers M, k, with 0<=k<=M, consider a representation of n as n = T(M) - T(M-k) = M + (M-1) + ... + (M-k+1), in which k is maximal, where T(r) = r*(r+1)/2 is the r-th triangular number. Then k = A109814(n), and M = A212652(n) = a(n) + (n/a(n) - 1)/2 is minimal.
Conjecture. For n, p, v, j natural numbers, the conditions on a(n) seem to be the following:
1. If n is an odd prime, then a(n) = 1.
2. If n is odd and composite, then
     a(n) = max(p : p | n, p <= sqrt(n), p is a prime).
3. If n is equal to a power of 2, then a(n) = n.
4. If n = 2^j*v, with v odd, v>1 and j>1, then a(n) = 2^j.
5. If n = 2*v, with v odd and composite, then
     a(n) = 2*p, where p is the least prime such that p | v.
6. If n = 2*p, for p an odd prime, then a(n) = 2.

T. D. Noe, Table of n, a(n) for n = 1..2048

Cf. A000217, A109814, A118235, A141419, A209260, A212652.

Table[d = Divisors[n]; mn = Infinity; best = 0; Do[q = i + (n/i - 1)/2; If[IntegerQ[q] && q < mn, mn = q; best = i], {i, d}]; best, {n, 100}] (* T. D. Noe, Feb 21 2013 *)

A346928 Irregular triangle read by rows; the n-th row contains, in ascending order, the distinct integers of the form n! / m! (with 1 <= m <= n) that do not appear in former rows.

Original entry on oeis.org

1, 2, 3, 6, 4, 12, 24, 5, 20, 60, 120, 30, 360, 720, 7, 42, 210, 840, 2520, 5040, 8, 56, 336, 1680, 6720, 20160, 40320, 9, 72, 504, 3024, 15120, 60480, 181440, 362880, 10, 90, 30240, 151200, 604800, 1814400, 3628800, 11, 110, 990, 7920, 55440, 332640, 1663200, 6652800, 19958400, 39916800

Offset: 1

Rémy Sigrist, Oct 16 2021

This sequence has similarities with A209260; here we consider quotients of factorial numbers, there differences of triangular numbers.
The n-th row is included in the n-th row of A068424 and has greatest term n!.
As a flat sequence, we have a permutation of the positive integers (any n > 0 appears among the first n rows, see A348401).
For any prime number p, the p-th row contains p-1 terms.

			Triangle begins:
    1;
    2;
    3, 6;
    4, 12, 24;
    5, 20, 60, 120;
    30, 360, 720;
    7, 42, 210, 840, 2520, 5040;
    8, 56, 336, 1680, 6720, 20160, 40320;
    9, 72, 504, 3024, 15120, 60480, 181440, 362880;
    10, 90, 30240, 151200, 604800, 1814400, 3628800;
    11, 110, 990, 7920, 55440, 332640, 1663200, 6652800, 19958400, 39916800;
    ...

Cf. A000142, A068424, A209260, A348401.

s=[]; for (n=1, 11, p=1; forstep (m=n, 1, -1, if (!setsearch(s, p*=m), s=setunion(s, [p]); print1 (p", "))))

from math import factorial
def auptor(rows):
    alst, aset = [1], {1}
    for n in range(2, rows+1):
        fn = factorial(n)
        for m in range(n-1, 0, -1):
            fm = factorial(m)
            q, r = divmod(fn, factorial(m))
            if r == 0 and q not in aset:
                alst.append(q); aset.add(q)
    return alst
print(auptor(11)) # Michael S. Branicky, Oct 17 2021

cp's OEIS Frontend

A141419 Triangle read by rows: T(n, k) = A000217(n) - A000217(n - k) with 1 <= k <= n.

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A212652 a(n) is the least positive integer M such that n = T(M) - T(k), for k an integer, 0 <= k <= M, where T(r) = A000217(r) is the r-th triangular number.

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A218621 a(n) = unique divisor d of n such that d + (n/d - 1)/2 is minimal and integral.

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A346928 Irregular triangle read by rows; the n-th row contains, in ascending order, the distinct integers of the form n! / m! (with 1 <= m <= n) that do not appear in former rows.

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