A144396 -id:A144396 - cp's OEIS frontend

A152579 a(n) = (10n+3)(10*n+17).

51, 351, 851, 1551, 2451, 3551, 4851, 6351, 8051, 9951, 12051, 14351, 16851, 19551, 22451, 25551, 28851, 32351, 36051, 39951, 44051, 48351, 52851, 57551, 62451, 67551, 72851, 78351, 84051, 89951, 96051, 102351, 108851, 115551, 122451, 129551, 136851, 144351, 152051, 159951

Offset: 0

Paul Curtz, Dec 08 2008

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A005563, A017270, A017305, A017353, A056220, A144396, A152161.

[(10*n+3)*(10*n+17): n in [0..40]]; // Vincenzo Librandi, Jul 28 2011

a(n)=100*n*(n+2)+51 \\ Charles R Greathouse IV, Jul 28 2011

a(n) = 100*n*(n+2) + 51 = 100*A005563(n) + 51 = 100*(n+1)^2 - 49 = A017270(n+1) - 49.
a(n) = 2*a(n-2) - a(n-2) + 200.
a(n) = 50*A056220(n+1) + 1.
a(n+1) - a(n) = 200*n + 300 = 100*A144396(n+1).
G.f.: (-51 - 198*x + 49*x^2)/(x-1)^3. - R. J. Mathar, Jul 01 2011
From Elmo R. Oliveira, Oct 27 2024: (Start)
E.g.f.: exp(x)*(51 + 300*x + 100*x^2).
a(n) = A017305(n)*A017353(n+1).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. (End)

A173154 a(n) = n^3/6 + 3n^2/4 + 7n/3 + 7/8 + (-1)^n/8.

Original entry on oeis.org

1, 4, 10, 19, 33, 52, 78, 111, 153, 204, 266, 339, 425, 524, 638, 767, 913, 1076, 1258, 1459, 1681, 1924, 2190, 2479, 2793, 3132, 3498, 3891, 4313, 4764, 5246, 5759, 6305, 6884, 7498, 8147, 8833, 9556, 10318, 11119, 11961, 12844, 13770, 14739, 15753, 16812, 17918, 19071, 20273, 21524

Offset: 0

Paul Curtz, Feb 11 2010

Generated by reading the table shown in A172002 down the diagonal starting at 1.

The inverse binomial transform yields 1, 3, 3, 0, 2, -4, 8, -16, 32, -64, 128, -256, 512, -1024, ... with a pattern of powers of 2.

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (3,-2,-2,3,-1).

[n^3/6 + 3*n^2/4 + 7*n/3 + 7/8 + (-1)^n/8: n in [0..50]]; // Vincenzo Librandi, Aug 05 2011

Table[n^3/6+(3n^2)/4+(7n)/3+7/8+(-1)^n/8,{n,0,50}] (* or *) LinearRecurrence[{3,-2,-2,3,-1},{1,4,10,19,33},50] (* Harvey P. Dale, Jan 04 2012 *)

G.f.: ( 1 + x - x^3 + x^4 ) / ( (1+x)*(x-1)^4 ).
a(n) = 3*a(n-1) - 2*a(n-2) - 2*a(n-3) + 3*a(n-4) - a(n-5).
a(n+4) - a(n) = 4*A152948(n+5) = 4*A089071(n+5).
First differences: a(n+1) - a(n) = A061925(n+2).
Second differences: a(n+2) - 2*a(n+1) + a(n) = n + 5/2 + (-1)^n/2 = 3, 3, 5, 5, 7, 7, 9, 9, ... , duplicated A144396.

A175717 First differences of A175628.

Original entry on oeis.org

0, 3, 2, 3, -5, 12, 6, 3, -22, 33, 10, 3, -33, 48, 14, 3, -74, 93, 18, 3, -85, 108, 22, 3, -156, 183, 26, 3, -161, 192, 30, 3, -268, 303, 34, 3, -261, 300, 38, 3, -410, 453, 42, 3, -385, 432, 46, 3, -582, 633, 50, 3, -533, 588, 54, 3, -784, 843, 58, 3, -705, 768, 62, 3, -1016, 1083, 66, 3, -901, 972, 70

Offset: 0

Paul Curtz, Dec 04 2010

G. C. Greubel, Table of n, a(n) for n = 0..5000
Index entries for linear recurrences with constant coefficients, signature (-1,-1,-1,-1,-1,-1,-1,2,2,2,2,2,2,2,2,-1,-1,-1,-1,-1,-1,-1,-1).

Cf. A131098, A144396, A175628.

R:= RealField(20);
b:= func< n | (n mod 2) eq 1 select (n-1)*(n+3)/4 else Round((n^2+4*n-12)*(37 +27*(-1)^(n/2) +6*Cos((n+2)*Pi(R)/4))/2^8) >;
[b(n+2) - b(n+1): n in [0..90]]; // G. C. Greubel, Sep 19 2018; Dec 04 2019

A175628 := proc(n) if type(n,'even') then nh := n/2 +1; 1/4-1/nh^2 ; numer(%) ; else nh := (n-1)/2 ; nh*(nh+2) ; end if; end proc:
A175717 := proc(n) A175628(n+2)-A175628(n+1) ; end proc:

LinearRecurrence[{-1,-1,-1,-1,-1,-1,-1,2,2,2,2,2,2,2,2,-1,-1,-1,-1,-1,-1,-1,-1},{0,3,2,3,-5,12,6,3,-22,33,10,3,-33,48,14,3,-74,93,18,3,-85,108,22},90] (* Harvey P. Dale, Apr 17 2015 *)
b[n_]:= If[OddQ[n], (n-1)*(n+3)/4, (n^2+4*n-12)*(37 +27*(-1)^(n/2) +6*Cos[(n+2)*Pi/4])/2^8]; Table[b[n+2] - b[n+1], {n, 0, 90}] (* G. C. Greubel, Dec 04 2019 *)

b(n) = if(n%2==1, (n-1)*(n+3)/4, round((n^2+4*n-12)*(37 +27*(-1)^(n/2) +6*cos((n+2)*Pi/4))/2^8) );
vector(91, n, b(n+1) - b(n) ) \\ G. C. Greubel, Sep 19 2018; Dec 04 2019

def b(n):
    if (mod(n,2)==1): return (n-1)*(n+3)/4
    else: return round((n^2+4*n-12)*(37 +27*(-1)^(n/2) +6*cos((n+2)*pi/4))/2^8)
[b(n+2) -b(n+1) for n in (0..90)] # G. C. Greubel, Dec 04 2019

a(2*n) + a(2*n+1) = A144396(n+1).
It seems that a(n) = 3 iff n in A131098.
a(n) = A175628(n+2) - A175628(n+1).
a(n)= -a(n-1) -a(n-2) -a(n-3) -a(n-4) -a(n-5) -a(n-6) -a(n-7) +2*a(n-8) +2*a(n-9) +2*a(n-10) +2*a(n-11) +2*a(n-12) +2*a(n-13) +2*a(n-14) +2*a(n-15) -a(n-16) -a(n-17) -a(n-18) -a(n-19) -a(n-20) -a(n-21) -a(n-22) -a(n-23). - R. J. Mathar, Dec 08 2010
From Luce ETIENNE, Dec 29 2019: (Start)
a(n) = 3*a(n-8) - 3*a(n-16) + a(n-24).
a(n) = (3*(264*m^7 - 6377*m^6 + 60963*m^5 - 293615*m^4 + 748881*m^3 - 962528*m^2 + 502812*m - 25200)*floor(n/8)^2 + 7*(136*m^7 - 3209*m^6 + 29731*m^5 - 137375*m^4 + 332209*m^3 - 400496*m^2 + 194844*m - 5040)*floor(n/8) + m*(472*m^6 - 11235*m^5 + 105049*m^4 - 488985*m^3 + 1181803*m^2 - 1389780*m + 617796))/5040, where m = n mod 8. (End)

A194130 a(n) = n!/gcd(n,3).

Original entry on oeis.org

1, 2, 2, 24, 120, 240, 5040, 40320, 120960, 3628800, 39916800, 159667200, 6227020800, 87178291200, 435891456000, 20922789888000, 355687428096000, 2134124568576000, 121645100408832000, 2432902008176640000, 17030314057236480000, 1124000727777607680000

Offset: 1

Paul Curtz, Aug 16 2011

nonn

Cf. A109007, A144396, A005563, A051925.

A194130 := proc(n)
        n!/igcd(n,3) ;
end proc:
seq(A194130(n),n=1..30) ;

Table[n!/GCD[n,3],{n,20}] (* Harvey P. Dale, Aug 08 2019 *)

Definition and offset corrected by R. J. Mathar, Aug 18 2011

A215415 a(2n) = n, a(4n+1) = 2n-1, a(4n+3) = 2*n+3.

Original entry on oeis.org

0, -1, 1, 3, 2, 1, 3, 5, 4, 3, 5, 7, 6, 5, 7, 9, 8, 7, 9, 11, 10, 9, 11, 13, 12, 11, 13, 15, 14, 13, 15, 17, 16, 15, 17, 19, 18, 17, 19, 21, 20, 19, 21, 23, 22, 21, 23, 25, 24, 23, 25, 27, 26, 25, 27, 29, 28, 27, 29, 31, 30, 29, 31, 33, 32, 31, 33, 35, 34, 33, 35, 37

Offset: 0

Paul Curtz, Aug 09 2012

a(n) and higher order differences in further rows:
0,  -1,  1,  3,  2,  1,
-1,  2,  2, -1, -1, -2,    A134430(n).
3,   0, -3,  0,  3,  0,
-3, -3,  3,  3, -3, -3,
0,   6,  0, -6,  0,  6,
6,  -6, -6,  6,  6, -6.
a(n) is the binomial transform of 0, -1, 3, -3, 0, 6, -12, 12, 0, -24, 48, -48, 0, 96..., essentially negated A134813.
By definition, all differences a(n+k)-a(n) are periodic sequences with period length 4. For k=1, 3 and 4 these are A134430, A021307 and A007395, for example.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,-2,2,-1).

Quadrisections: A005843, A060747, A005408, A144396.

Flatten[Table[{2n, 2n - 1, 2n + 1, 2n + 3}, {n, 0, 19}]] (* Alonso del Arte, Aug 09 2012 *)

a(n) = ((-3*I)*((-I)^n-I^n)+2*n)/4 \\ Colin Barker, Oct 19 2015

concat(0, Vec(-x*(1-3*x+x^2)/((x^2+1)*(x-1)^2) + O(x^100))) \\ Colin Barker, Oct 19 2015

a(2*n) = n, a(2*n+1) = A097062(n+1).
a(n) = (A214297(n+1) - A214297(n-1))/2.
a(3*n) =3*A004525(n).
a(n) = +2*a(n-1) -2*a(n-2) +2*a(n-3) -a(n-4).
G.f. -x*(1-3*x+x^2) / ( (x^2+1)*(x-1)^2 ). - R. J. Mathar, Aug 11 2012
a(n) = ((-3*I)*((-I)^n-I^n)+2*n)/4. - Colin Barker, Oct 19 2015

A158897 The elements of A059100 at indices of triangular numbers, padded with zeros.

Original entry on oeis.org

6, 0, 11, 0, 0, 18, 0, 0, 0, 27, 0, 0, 0, 0, 38, 0, 0, 0, 0, 0, 51, 0, 0, 0, 0, 0, 0, 66, 0, 0, 0, 0, 0, 0, 0, 83, 0, 0, 0, 0, 0, 0, 0, 0, 102, 0, 0, 0, 0, 0, 0, 0, 0, 0, 123, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 146, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 171, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 198, 0, 0, 0, 0, 0, 0

Offset: 1

Paul Curtz, Mar 29 2009

nonn

Cf. A144396.

a(t)=A059100(i+1) if t=A000217(i), else a(t)=0.

a(n)=A141620(n-1)+A158405(n+1).

Edited and extended by R. J. Mathar, Apr 04 2009

A182397 Numerators in triangle that leads to the (first) Bernoulli numbers A027641/A027642.

Original entry on oeis.org

1, 1, -3, 1, -5, 5, 1, -7, 25, -5, 1, -9, 23, -35, 49, 1, -11, 73, -27, 112, -49, 1, -13, 53, -77, 629, -91, 58, 1, -15, 145, -130, 1399, -451, 753, -58, 1, -17, 95, -135, 2699, -2301, 8573, -869, 341, 1, -19, 241

Offset: 0

Paul Curtz, Apr 27 2012

In A190339 we saw that (the second Bernoulli numbers) A164555/A027642 is an eigensequence (its inverse binomial transform is the sequence signed) of the second kind, see A192456/A191302. We consider this array preceded by 1 for the second row, by 1, -3/2, for the third one; 1 is chosen and is followed by the differences of successive rows.
Hence
                    1    1/2   1/6      0
            1    -1/2   -1/3  -1/6  -1/30
      1   -3/2    1/6    1/6  2/15   1/15
  1 -5/2   5/3      0  -1/30 -1/15 -8/105.
The second row is A051716/A051717.
The (reduced) triangle before the square array (T(n,m) in A190339) is a(n)/b(n)=
B(0)=    1 = 1                 Redbernou1li
B(1)= -1/2 = 1  -3/2
B(2)=  1/6 = 1  -5/2  5/3
B(3)=    0 = 1  -7/2 25/6  -5/3
B(4)=-1/30 = 1  -9/2 23/3 -35/6  49/30
B(5)=    0 = 1 -11/2 73/6 -27/2 112/15 -49/30.
For the main diagonal, see A165142.
Denominator b(n) will be submitted.
This transform is valuable for every eigensequence of the second kind. For instance Leibniz's 1/n (A003506).
With increasing exponents for coefficients, polynomials CB(n,x) create Redbernou1li. See the formula.
Triangle Bernou1li for A027641/A027642 with the same denominator A080326 for every column is
1
1  -3/2
1  -5/2 10/6
1  -7/2 25/6 -10/6
1  -9/2 46/6 -35/6  49/30
1 -11/2 73/6 -81/6 224/30 -49/30.
For numerator by columns,see A000012, -A144396, A100536, Q(n)=n*(2*n^2+9*n+9)/2 , new.
Triangle Checkbernou1 with the same denominator A080326 for every row is
1/1
(2    -3)/2
(6   -15  +10)/6
(6   -21  +25  -10)/6
(30 -135 +230 -175  +49)/30
(30 -165 +365 -405 +224 -49)/30;
Hence for numerator: 1, 2-3, 16-15, 31-31, 309-310, 619-619, 8171-8166.
Absolute sum: 1, 5, 31, 62, 619, 1238, 17337. Reduced division by A080326:
1, 5/2, 31/6, 31/3, 619/30, 619/15, 5779/70, = A172030(n+1)/A172031(n+1).

Cf. A028246 (Worpitzky), A085737/A085738 (Conway-Sloane), A051714/A051715 (Akiyama-Tanigawa), A192456/A191302 for other triangles that lead to the Bernoulli numbers.

CB(0,x) = 1,
CB(1,x) = 1 - 3*x/2,
CB(n,x) = (1-x)*CB(n-1,x) + B(n)*x^n , n > 1.

A256455 Numbers that appear at least once in a Pythagorean triple (a, b, b+1).

Original entry on oeis.org

3, 4, 5, 7, 9, 11, 12, 13, 15, 17, 19, 21, 23, 24, 25, 27, 29, 31, 33, 35, 37, 39, 40, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 60, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 84, 85, 87, 89, 91, 93, 95, 97, 99, 101, 103, 105, 107, 109, 111, 112, 113

Offset: 1

J. Lowell, Mar 29 2015

nonn

Includes all odd numbers >= 3 because every odd number a has a Pythagorean triple (a, b, b+1).

Union of A144396 and A046092 (except for 0). - Robert Israel, Mar 29 2015

			12 qualifies because it's part of (5, 12, 13). 8 doesn't qualify because no Pythagorean triple of the form (a, b, b+1) has 8 in it; in every triple of this kind, b is the only even number, and a in the triple (a, 8, 9) would be the square root of 17, which is not an integer.

Cf. A144396 (the values of a), A046092 (the values of b), A001844 (the values of b+1).

N:= 500: # to get all terms up to N
sort([seq(2*i+1, i=1 .. floor((N-1)/2)), seq(2*j*(j+1), j = 1 .. floor((sqrt(1+2*N)-1)/2))]); # Robert Israel, Mar 29 2015

A370481 a(0) = 33. a(n) = 3a(n-1) + 2n + 1 for n >= 1.

Original entry on oeis.org

33, 102, 311, 940, 2829, 8498, 25507, 76536, 229625, 688894, 2066703, 6200132, 18600421, 55801290, 167403899, 502211728, 1506635217, 4519905686, 13559717095, 40679151324, 122037454013, 366112362082, 1098337086291, 3295011258920, 9885033776809, 29655101330478

Offset: 0

Paul Curtz, Mar 31 2024

Last digit is of period 10: repeat [3, 2, 1, 0, 9, 8, 7, 6, 5, 4].

			a(1) = 3*33 + 3 = 102, a(2) = 3*102 + 5 = 311.

Index entries for linear recurrences with constant coefficients, signature (5,-7,3).

Cf. A144396, A237930.

LinearRecurrence[{5, -7, 3}, {33, 102, 311}, 26] (* Amiram Eldar, Apr 01 2024 *)

a(n) = 4*a(n-1) - 3*a(n-2) + 2 with a(0) = 33 and a(1) = 102 for n >= 2.
a(n) = 10*A237930(n) + 3 - n.
a(n) = 35*3^n - n - 2. - Hugo Pfoertner, Mar 31 2024

A386319 Triangle read by rows where row n is the start, corner and end vertex numbers of a triangular spiral with n sides on a triangular grid, starting from 1 and working inwards (0 <= k <= n).

Original entry on oeis.org

1, 1, 1, 1, 2, 3, 1, 3, 5, 6, 1, 4, 7, 9, 10, 1, 5, 9, 12, 14, 15, 1, 6, 11, 15, 18, 20, 21, 1, 7, 13, 18, 22, 25, 27, 28, 1, 8, 15, 21, 26, 30, 33, 35, 36, 1, 9, 17, 24, 30, 35, 39, 42, 44, 45, 1, 10, 19, 27, 34, 40, 45, 49, 52, 54, 55, 1, 11, 21, 30, 38, 45, 51, 56, 60, 63, 65, 66, 1, 12, 23, 33, 42, 50, 57, 63, 68, 72, 75, 77, 78

Offset: 0

Binay Krishna Maity, Jul 18 2025

The first 2 sides are length n-1 so that T(n,1) = 1 + (n-1) and T(n,2) = 1 + 2*(n-1) and then the side lengths decrease by 1 each time as it spirals in (ending at triangular number A000217(n) when n>=1).
These sides mesh to fill the triangle as they go inwards, and can also be thought of going outwards tracing out the sides of the triangle.
The resulting vertex numbers are 1 together with row n of A141419.
Row n=1 is taken as a side of length 0 so the start and end numbers are both 1 (which is not really a spiral but is consistent with the formula and two points 1,2 would be even less like a triangle filled by a spiral).

			Triangle begins:
--------------------------------------
   n\k  0   1   2   3   4   5   6   7
--------------------------------------
   0|   1;
   1|   1,  1;
   2|   1,  2,  3;
   3|   1,  3,  5,  6;
   4|   1,  4,  7,  9, 10;
   5|   1,  5,  9, 12, 14, 15;
   6|   1,  6, 11, 15, 18, 20, 21;
   7|   1,  7, 13, 18, 22, 25, 27, 28;
  ...
For n = 2 the spiral is 2 sides of length 1 so row [1, 2, 3],
   1 --- 2
       /
     3
For n = 4 the spiral is:
   1  2  3  4
    9  10  5
      8  6
        7
The start, corner and end vertices are [1, 4, 7, 9, 10].

Columns: A000012 (k=0), A000027 (k=1), A144396 (k=3).
Cf. A179865(n+1) (main diagonal), A056520 (row sums).
Cf. A000217, A141419.

T[n_,k_]:=If[k==0,1,k(2n-k+1)/2];Table[T[n,k],{n,0,12},{k,0,n}]//Flatten (* James C. McMahon, Jul 31 2025 *)

T(n,0) = 1.

T(n,k) = k*(2*n - k + 1)/2 for k >= 1.

cp's OEIS Frontend

A152579 a(n) = (10*n+3)*(10*n+17).

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A173154 a(n) = n^3/6 + 3*n^2/4 + 7*n/3 + 7/8 + (-1)^n/8.

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A175717 First differences of A175628.

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A194130 a(n) = n!/gcd(n,3).

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A215415 a(2*n) = n, a(4*n+1) = 2*n-1, a(4*n+3) = 2*n+3.

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A158897 The elements of A059100 at indices of triangular numbers, padded with zeros.

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A182397 Numerators in triangle that leads to the (first) Bernoulli numbers A027641/A027642.

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A256455 Numbers that appear at least once in a Pythagorean triple (a, b, b+1).

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A370481 a(0) = 33. a(n) = 3*a(n-1) + 2*n + 1 for n >= 1.

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A152579 a(n) = (10n+3)(10*n+17).

A173154 a(n) = n^3/6 + 3n^2/4 + 7n/3 + 7/8 + (-1)^n/8.

A215415 a(2n) = n, a(4n+1) = 2n-1, a(4n+3) = 2*n+3.

A370481 a(0) = 33. a(n) = 3a(n-1) + 2n + 1 for n >= 1.