A193910 -id:A193910 - cp's OEIS frontend

A274406 Numbers m such that 9 divides m*(m + 1).

0, 8, 9, 17, 18, 26, 27, 35, 36, 44, 45, 53, 54, 62, 63, 71, 72, 80, 81, 89, 90, 98, 99, 107, 108, 116, 117, 125, 126, 134, 135, 143, 144, 152, 153, 161, 162, 170, 171, 179, 180, 188, 189, 197, 198, 206, 207, 215, 216, 224, 225, 233, 234, 242, 243, 251, 252, 260, 261, 269

Offset: 1

Bruno Berselli, Jun 20 2016

Equivalently, numbers congruent to 0 or 8 mod 9.

Terms of A007494 with indices in A047264. Also, terms of A060464 with indices in A047335.

Bruno Berselli, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A008591 (first bisection), A010689 (first differences), A017257 (second bisection).
Cf. similar sequences in which m*(m+1) is divisible by k: A014601 (k=4), A047208 (k=5), A007494 (k=3 and 6), A047335 (k=7), A047521 (k=8), this sequence (k=9).
Cf. A301451: numbers congruent to {1, 7} mod 9; A193910: numbers congruent to {2, 6} mod 9.

[n: n in [0..300] | IsDivisibleBy(n*(n+1),9)];

Select[Range[0, 300], Divisible[# (# + 1), 9] &]

for(n=0, 300, if(n*(n+1)%9==0, print1(n", ")))

[n for n in range(300) if 9.divides(n*(n+1))]

G.f.: x^2*(8 + x)/((1 + x)*(1 - x)^2).
a(n) = (18*n + 7*(-1)^n - 11)/4. Therefore: a(2*m) = 9*m-1, a(2*m+1) = 9*m. It follows that a(j)+a(k) and a(j)*a(k) belong to the sequence if j and k are not both even.
a(n) = -A090570(-n+2).
a(n) = a(n-1) + a(n-2) - a(n-3).
a(2*r+1) + a(2*r+s+1) = a(4*r+s+1) and a(2*r) + a(2*r+2*s+1) = a(4*r+2*s). A particular case provided by these identities: a(n) = a(n - 2*floor(n/6)) + a(2*floor(n/6) + 1).
E.g.f.: 1 + ((9*x - 2)*cosh(x) + 9*(x - 1)*sinh(x))/2. - Stefano Spezia, Apr 24 2021

A301451 Numbers congruent to {1, 7} mod 9.

Original entry on oeis.org

1, 7, 10, 16, 19, 25, 28, 34, 37, 43, 46, 52, 55, 61, 64, 70, 73, 79, 82, 88, 91, 97, 100, 106, 109, 115, 118, 124, 127, 133, 136, 142, 145, 151, 154, 160, 163, 169, 172, 178, 181, 187, 190, 196, 199, 205, 208, 214, 217, 223, 226, 232, 235, 241, 244, 250, 253, 259, 262, 268

Offset: 1

Bruno Berselli, Mar 21 2018

First bisection of A056991, second bisection of A242660.

The squares of the terms of A174396 are the squares of this sequence.

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

Cf. A274406: numbers congruent to {0, 8} mod 9.
Cf. A193910: numbers congruent to {2, 6} mod 9.
Subsequence of A016777, A026225, A029739, A033627, A047236, A047259, A055047, A055054, A056991, A153053, A187318, A242660.

a := [1,7,10];; for n in [4..60] do a[n] := a[n-1] + a[n-2] - a[n-3]; od; a;

Magma
```
&cat [[9*n+1, 9*n+7]: n in [0..40]];
```

Table[2 (2 n - 1) + (2 n - 3 (1 - (-1)^n))/4, {n, 1, 60}]
{#+1,#+7}&/@(9*Range[0,30])//Flatten (* or *) LinearRecurrence[{1,1,-1},{1,7,10},60] (* Harvey P. Dale, Nov 08 2020 *)

Vec(x*(1 + 6*x + 2*x^2) / ((1 - x)^2*(1 + x)) + O(x^60)) \\ Colin Barker, Mar 22 2018

[2*(2*n-1)+(2*n-3*(1-(-1)**n))/4 for n in range(1,70)]

[n for n in (1..300) if n % 9 in (1,7)]

O.g.f.: x*(1 + 6*x + 2*x^2)/((1 + x)*(1 - x)^2).
E.g.f.: (3 + 8*exp(x) - 11*exp(2*x) + 18*x*exp(2*x))*exp(-x)/4.
a(n) = a(n-1) + a(n-2) - a(n-3).
a(n) = 2*(2*n - 1) + (2*n - 3*(1 - (-1)^n))/4. Therefore, for n even a(n) = (9*n - 4)/2, otherwise a(n) = (9*n - 7)/2.
a(2n+1) = A017173(n). a(2n) = A017245(n-1). - R. J. Mathar, Feb 28 2019

A219795 Sum of the absolute values of the antidiagonals of the triangle A135929(n) companion. See the comment.

Original entry on oeis.org

2, 2, 2, 2, 3, 3, 5, 7, 10, 11, 16, 23, 33, 44, 58, 81, 114, 158, 212, 293, 407, 565, 777, 1064, 1471, 2036, 2813, 3863, 5334, 7370, 10183, 14046, 19356, 26726, 36909, 50955, 70251, 96977, 133886, 184841, 255092

Offset: 0

Paul Curtz, Nov 28 2012

nonn

The companion to A135929(n) is the triangle
2;
2, 0;
2, 0,  1;
2, 0, -1, 0;
2, 0, -3, 0, -1;
2, 0, -5, 0,  0, 0;
2, 0, -7, 0,  3, 0, 1;
2, 0, -9, 0,  8, 0, 1, 0;
(A192011(n) beginning with 2 instead of -1).
Consider a(1),a(5),a(10),a(14), that is, a(A193910(n) -1).
a(1)+a(4)-a(5) = 2, a(5)+a(8)-a(9) = 2, a(10)+a(13)-a(14) = 2, a(14)+a(17)-a(18) = 4, a(19)+a(22)-a(23) = 6, a(23)+a(26)-a(27) = 14, yields 2,2,2,4,6,14,24,60,... = 2*A047749(n) or 2, followed with A116637(n+1).

			a(0)=2, a(1)=2, a(2)=2+0, a(3)=2+0, a(4)=2+0+1, a(5)=2+0+1.

A219795 := proc(n)
    if n=0 then
        2;
    else
        add(abs(A192011(n-k,k)),k=0..floor(n/2)) ;
    end if;
end proc: # R. J. Mathar, Jan 06 2013

a(n) = sum abs ( [k=0..floor(n/2)] A192011(n-k,k) ), a(0)=2.

a(24)-a(40) from Jean-Francois Alcover, Nov 28 2012

A193879 Different leap years in the Gregorian and the revised Julian calendars.

Original entry on oeis.org

0, 2, 4, 6, 8, 11, 12, 15, 16, 28, 29, 32, 33, 36, 38, 40, 42, 44, 47, 48, 51, 52, 64, 65, 68, 69, 72, 74, 76, 78, 80, 83, 84, 87, 88, 100, 101, 104, 105, 108, 110, 112, 114, 116, 119, 120, 123, 124, 136, 137, 140, 141, 144, 146, 148, 150, 152, 155, 156, 159, 160

Offset: 1

Frank Ellermann, Aug 07 2011

Terms divided by 100, e.g., 28 indicates year 2800, which is a Gregorian leap year, but not a revised Julian leap year. Values below 28 are "proleptic" (only based on the formula).

			28 mod 9 is not 2 or 6, but 28 mod 4 is 0: 2800 is a Gregorian leap year.
29 mod 9 is 2, but 29 mod 4 is not 0: 2900 is a revised Julian leap year.

M. Milankovitch, Das Ende des julianischen Kalenders und der neue Kalender der orientalischen Kirchen, Astronomische Nachrichten, volume 220 (1924), pages 379-384.
Claus Tøndering, Frequently Asked Questions about Calendars, Don't the Greeks do it differently?
Wikipedia, Revised Julian calendar

A008586 enumerates "Gregorian leap centuries" (N // 4 = 0).

A193910 enumerates "revised Julian leap centuries".

( N // 9 = 2 | N // 9 = 6 ) <> ( N // 4 = 0 )

A144652 Triangle, read by rows, where T(m,n) = floor((2mn+m+n)/2) with m >= n >= 1.

Original entry on oeis.org

2, 3, 6, 5, 8, 12, 6, 11, 15, 20, 8, 13, 19, 24, 30, 9, 16, 22, 29, 35, 42, 11, 18, 26, 33, 41, 48, 56, 12, 21, 29, 38, 46, 55, 63, 72, 14, 23, 33, 42, 52, 61, 71, 80, 90, 15, 26, 36, 47, 57, 68, 78, 89, 99, 110, 17, 28, 40, 51, 63, 74, 86, 97, 109, 120, 132, 18, 31, 43, 56, 68

Offset: 1

Vincenzo Librandi, Jan 27 2009

From Vincenzo Librandi, Nov 16 2012: (Start)
First column:  A007494(n+1);
second column: A047219(n+2);
third column:  A047383(n+3);
fourth column: A193910(n+4).
Conjecture: If h does not belong to the sequence, then 4*h+1 is prime. (End)

			Triangle begins:
2;
3,  6;
5,  8,  12;
6,  11, 15, 20;
8,  13, 19, 24, 30;
9,  16, 22, 29, 35, 42;
11, 18, 26, 33, 41, 48, 56; etc.

Vincenzo Librandi, Rows n = 1..100, flattened

Cf. A007494, A047219, A047383, A193910.

[Floor((2*n*k+n+k)/2): k in [1..n], n in [1..11]]; // Vincenzo Librandi, Nov 16 2012

Flatten[Table[Floor[(2*n*m + m + n)/2], {n, 1, 20}, {m, n}]] (* Vincenzo Librandi, Nov 16 2012 *)

Definition edited (specifying m >= n >= 1), and terms recomputed to match definition, as was done with the similar sequence A140869, by Jon E. Schoenfield, Jun 24 2010

A215414 Unix epoch timestamp for start of year, beginning with 1970.

Original entry on oeis.org

0, 31536000, 63072000, 94694400, 126230400, 157766400, 189302400, 220924800, 252460800, 283996800, 315532800, 347155200, 378691200, 410227200, 441763200, 473385600, 504921600, 536457600, 567993600, 599616000, 631152000, 662688000, 694224000, 725846400, 757382400

Offset: 1

Kyle Stern, Aug 09 2012

This is based on a naive multiplication of A033172 with a fixed number of seconds per year, 24*3600 = 86400. It ignores that leap years are not regularly occurring after 4 years (but after 400 years, note the formula that relates a(n+4) to a(n) and also the simple Mma implementation), ignores leap seconds, and any other influences that align the slowing down of the Earth rotation in an astronomical fixed coordinate system measured relative to atomic clocks. In summary, the use of "year" in the definition is not commensurate with years in standard astronomical or earth observational terms. - R. J. Mathar, Aug 21 2012

G. C. Greubel, Table of n, a(n) for n = 1..1000
Wikipedia, Unix time
Wikipedia, Leap second
Wikipedia, Revised Julian Calendar
Wikipedia, International Earth Rotation and Reference Systems Service

Cf. A033172, A193910.

lst = {}; t = 86400; Do[e = t*(365*(n - 1) + Ceiling[n/4]); If[! Mod[n, 4] == 0, e = e - t]; AppendTo[lst, e], {n, 25}]; lst (* Arkadiusz Wesolowski, Aug 20 2012 *)
CoefficientList[Series[86400*(365*x + 365*x^2 + 366*x^3 + 365*x^4)/((x - 1)^2*(1 +x +x^2 +x^3)), {x,0,50}], x] (* G. C. Greubel, Feb 26 2017 *)

x='x+O('x^50); Vec(86400*(365*x +365*x^2 +366*x^3 +365*x^4)/((1-x)^2*(1+x+x^2+x^3))) \\ G. C. Greubel, Feb 26 2017

From Alexander R. Povolotsky, Aug 20 2012: (Start)
a(n) = 10800*(2922*n + (-1)^n + (1+i)*(-i)^n + (1-i)*i^n - 2923).
a(n+4) = a(n) + 126230400.
G.f.: 86400*(365*x +365*x^2 +366*x^3 +365*x^4)/((1-x)^2*(1+x+x^2+x^3)). (End)

a(11)-a(25) from Arkadiusz Wesolowski, Aug 20 2012

cp's OEIS Frontend

A274406 Numbers m such that 9 divides m*(m + 1).

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A301451 Numbers congruent to {1, 7} mod 9.

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A219795 Sum of the absolute values of the antidiagonals of the triangle A135929(n) companion. See the comment.

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Maple

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A193879 Different leap years in the Gregorian and the revised Julian calendars.

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A144652 Triangle, read by rows, where T(m,n) = floor((2mn+m+n)/2) with m >= n >= 1.

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Magma

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A215414 Unix epoch timestamp for start of year, beginning with 1970.

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Mathematica

PARI

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