A185787 - OEIS

A185787

Sum of first k numbers in column k of the natural number array A000027; by antidiagonals.

1, 7, 25, 62, 125, 221, 357, 540, 777, 1075, 1441, 1882, 2405, 3017, 3725, 4536, 5457, 6495, 7657, 8950, 10381, 11957, 13685, 15572, 17625, 19851, 22257, 24850, 27637, 30625, 33821, 37232, 40865, 44727, 48825, 53166, 57757, 62605, 67717, 73100, 78761, 84707, 90945, 97482, 104325, 111481, 118957, 126760, 134897, 143375

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OFFSET

1,2

COMMENTS

This is one of many interesting sequences and arrays that stem from the natural number array A000027, of which a northwest corner is as follows:

1....2.....4.....7...11...16...22...29...

3....5.....8....12...17...23...30...38...

6....9....13....18...24...31...39...48...

10...14...19....25...32...40...49...59...

15...20...26....33...41...50...60...71...

21...27...34....42...51...61...72...84...

28...35...43....52...62...73...85...98...

Blocking out all terms below the main diagonal leaves columns whose sums comprise A185787. Deleting the main diagonal and then summing give A185787. Analogous treatments to the left of the main diagonal give A100182 and A101165. Further sequences obtained directly from this array are easily obtained using the following formula for the array: T(n,k)=n+(n+k-2)(n+k-1)/2.

Examples:

row 1: A000124

row 2: A022856

row 3: A016028

row 4: A145018

row 5: A077169

col 1: A000217

col 2: A000096

col 3: A034856

col 4: A055998

col 5: A046691

col 6: A052905

col 7: A055999

diag. (1,5,...) ...... A001844

diag. (2,8,...) ...... A001105

diag. (4,12,...)...... A046092

diag. (7,17,...)...... A056220

diag. (11,23,...) .... A132209

diag. (16,30,...) .... A054000

diag. (22,38,...) .... A090288

diag. (3,9,...) ...... A058331

diag. (6,14,...) ..... A051890

diag. (10,20,...) .... A005893

diag. (15,27,...) .... A097080

diag. (21,35,...) .... A093328

Antidiagonal sums: (1,5,15,34,...)=A006003=partial sums of A002817.

Let S(n,k) denote the n-th partial sum of column k. Then

S(n,k)=n*(n^2+3k*n+3*k^2-6*k+5)/6.

S(n,1)=n(n+1)(n+2)/6

S(n,2)=n(n+1)(n+5)/6

S(n,3)=n(n+2)(n+7)/6

S(n,4)=n(n^2+12n+29)/6

S(n,5)=n(n+5)(n+10)/6

S(n,6)=n(n+7)(n+11)/6

S(n,7)=n(n+10)(n+11)/6

Weight array of T: A144112

Accumulation array of T: A185506

Second rectangular sum array of T: A185507

Third rectangular sum array of T: A185508

Fourth rectangular sum array of T: A185509

From Clark Kimberling, Oct 13 2025: (Start)

Continuing with the natural number array T, for n>=1,

* number of the row containing n: A002260(n);

* number of the column containing n: A004736(n);

* number of the antidiagonal containing n: A002024(n);

* self-convolution of n-th antidiagonal: A387429(n);

* number of squares in n-th antidiagonal: A361433(n).

* number of primes in n-th antidiagonal: A066888(n);

* number of odd numbers in n-th antidiagonal: A004525(n);

* number of even numbers in n-th antidiagonal: A004524(n) for n>=2;

* Every multisection of every row, every column, and every diagonal is linearly recurrent with signature (3,-3, 1). Indeed, every "linear staircase sequence", T(n+h*u,k+h*v), for fixed n,k,u,v, with h=0..oo, is linearly recurrent with signature (3,-3,1).

Let F(n) be the set of the first n numbers in row n together with the first n numbers in column n, so that F(1) = {1}, F(2) = {2,3,5}, F(3) = {4,6,8,9,13}, etc. Then n is in F(A051125(n)).

(End)

LINKS

Vincenzo Librandi, Table of n, a(n) for n = 1..1000

Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

FORMULA

a(n) = n*(7*n^2-6*n+5)/6.

G.f.: x*(3*x^2+3*x+1)/(1-x)^4. - Vincenzo Librandi, Jul 04 2012

MATHEMATICA

f[n_, k_]:=n+(n+k-2)(n+k-1)/2;

s[k_]:=Sum[f[n, k], {n, 1, k}];

Factor[s[k]]

Table[s[k], {k, 1, 70}] (* A185787 *)