A011900 -id:A011900

A113225

a(2n) = A011900(n), a(2n+1) = A001109(n+1).

+20
3

1, 1, 3, 6, 15, 35, 85, 204, 493, 1189, 2871, 6930, 16731, 40391, 97513, 235416, 568345, 1372105, 3312555, 7997214, 19306983, 46611179, 112529341, 271669860, 655869061, 1583407981, 3822685023, 9228778026, 22280241075, 53789260175

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A011906

If b(n) is A011900(n) and c(n) is A001109(n), then a(n) = b(n)*c(n) = b(n) + (b(n)+1) + (b(n)+2) + ... + c(n).

+20
2

1, 18, 525, 17340, 586177, 19896030, 675781821, 22956120408, 779829016225, 26491211221770, 899921240562957, 30570830315362260, 1038508305678375841, 35278711540581704598, 1198437683944896688125, 40711602541832856049200, 1382996048733983114022337

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MAPLE

A011900 := proc(n) coeftayl( (1-4*x+x^2)/((1-x)*(1-6*x+x^2)), x=0, n) ; end proc: A001109 := proc(n) coeftayl( x/(1-6*x+x^2), x=0, n) ; end proc: A011906 := proc(n) A001109(n)*A011900(n-1) ; end proc: seq(A011906(n), n=1..30) ; # R. J. Mathar, Apr 15 2010

CROSSREFS

Cf. A001109, A011900, A029547, A091761.

A001653

Numbers k such that 2*k^2 - 1 is a square.
(Formerly M3955 N1630)

+10
205

1, 5, 29, 169, 985, 5741, 33461, 195025, 1136689, 6625109, 38613965, 225058681, 1311738121, 7645370045, 44560482149, 259717522849, 1513744654945, 8822750406821, 51422757785981, 299713796309065, 1746860020068409, 10181446324101389, 59341817924539925

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FORMULA

For n < k, a(n)*A001541(k) = A011900(n+k)+A053141(k-n-1); e.g., 5*99 = 495 = 493+2. For n >= k, a(n)*A001541(k) = A011900(n+k)+A053141(n-k); e.g., 29*3 = 87 = 85+2. - Charlie Marion, Oct 18 2004

A001109

a(n)^2 is a triangular number: a(n) = 6*a(n-1) - a(n-2) with a(0)=0, a(1)=1.
(Formerly M4217 N1760)

+10
193

0, 1, 6, 35, 204, 1189, 6930, 40391, 235416, 1372105, 7997214, 46611179, 271669860, 1583407981, 9228778026, 53789260175, 313506783024, 1827251437969, 10650001844790, 62072759630771, 361786555939836, 2108646576008245, 12290092900109634, 71631910824649559, 417501372047787720

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FORMULA

a(n) = A002315(m)*A011900(n-m-1) + A001653(m)*A001652(n-m-1) - a(m) = A002315(m)*A053141(n-m-1) + A001653(m)*A046090(n-m-1) + a(m) with m < n; otherwise a(n) = A002315(m)*A053141(m-n) - A001653(m)*A011900(m-n) + a(m) = A002315(m)*A053141(m-n) - A001653(m)*A046090(m-n) - a(m) = (A002315(n) - A001653(n))/2. - Kenneth J Ramsey, Oct 12 2011

A001541

a(0) = 1, a(1) = 3; for n > 1, a(n) = 6*a(n-1) - a(n-2).
(Formerly M3037 N1231)

+10
116

1, 3, 17, 99, 577, 3363, 19601, 114243, 665857, 3880899, 22619537, 131836323, 768398401, 4478554083, 26102926097, 152139002499, 886731088897, 5168247530883, 30122754096401, 175568277047523, 1023286908188737, 5964153172084899, 34761632124320657

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FORMULA

For n > k, a(n)*A001653(k) = A011900(n+k) + A053141(n-k-1). For n <= k, a(n)*A001653(k) = A011900(n+k) + A053141(k-n). - Charlie Marion, Oct 18 2004

A001110

Square triangular numbers: numbers that are both triangular and square.
(Formerly M5259 N2291)

+10
83

0, 1, 36, 1225, 41616, 1413721, 48024900, 1631432881, 55420693056, 1882672131025, 63955431761796, 2172602007770041, 73804512832419600, 2507180834294496361, 85170343853180456676, 2893284510173841030625, 98286503002057414584576, 3338847817559778254844961, 113422539294030403250144100

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FORMULA

a(n) = (A011900(n) + A001652(n))^2; see the link about the generalized proof of square triangular numbers. - Kenneth J Ramsey, Oct 10 2011

A053141

a(0)=0, a(1)=2 then a(n) = a(n-2) + 2*sqrt(8*a(n-1)^2 + 8*a(n-1) + 1).

+10
47

0, 2, 14, 84, 492, 2870, 16730, 97512, 568344, 3312554, 19306982, 112529340, 655869060, 3822685022, 22280241074, 129858761424, 756872327472, 4411375203410, 25711378892990, 149856898154532, 873430010034204, 5090723162050694, 29670908962269962, 172934730611569080

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COMMENTS

The values x(n)=A001652(n), y(n)=A046090(n) and z(n)=A001653(n) form a nearly isosceles Pythagorean triple since y(n)=x(n)+1 and x(n)^2 + y(n)^2 = z(n)^2; e.g., for n=2, 20^2 + 21^2 = 29^2. In a similar fashion, if we define b(n)=A011900(n) and c(n)=A001652(n), a(n), b(n) and c(n) form a nearly isosceles anti-Pythagorean triple since b(n)=a(n)+1 and a(n)^2 + b(n)^2 = c(n)^2 + c(n) + 1; i.e., the value a(n)^2 + b(n)^2 lies almost exactly between two perfect squares; e.g., 2^2 + 3^2 = 13 = 4^2 - 3 = 3^2 + 4; 14^2 + 15^2 = 421 = 21^2 - 20 = 20^2 + 21. - Charlie Marion, Jun 12 2009

FORMULA

a(n) = (A001653(n)-1)/2 = 2*A053142(n) = A011900(n)-1. [Corrected by Pontus von Brömssen, Sep 11 2024]

For n>k, a(n-k-1) = A001541(n)*A001653(k)-A011900(n+k); e.g., 2 = 99*5 - 493.

For n<=k, a(k-n) = A001541(n)*A001653(k) - A011900(n+k); e.g., 2 = 3*29 - 85 + 2. (End)

CROSSREFS

Cf. A011900, A029549, A053142, A075528, A103200.

A029549

a(n + 3) = 35*a(n + 2) - 35*a(n + 1) + a(n), with a(0) = 0, a(1) = 6, a(2) = 210.

+10
39

0, 6, 210, 7140, 242556, 8239770, 279909630, 9508687656, 323015470680, 10973017315470, 372759573255306, 12662852473364940, 430164224521152660, 14612920781245825506, 496409142337836914550, 16863297918705209269200, 572855720093639278238256

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FORMULA

a(n) = A002378(A011900(n)-1) = A002378(A053141(n)). - Pontus von Brömssen, Sep 11 2024

CROSSREFS

Cf. A000217, A001109, A001652, A002315, A002378, A011900, A029546, A046090, A046729, A053141, A084159, A096979, A157259, A165518.

A046090

Consider all Pythagorean triples (X,X+1,Z) ordered by increasing Z; sequence gives X+1 values.

+10
38

1, 4, 21, 120, 697, 4060, 23661, 137904, 803761, 4684660, 27304197, 159140520, 927538921, 5406093004, 31509019101, 183648021600, 1070379110497, 6238626641380, 36361380737781, 211929657785304, 1235216565974041, 7199369738058940, 41961001862379597, 244566641436218640

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COMMENTS

Solution to a*(a-1) = 2b*(b-1) in natural numbers: a = a(n), b = b(n) = A011900(n).

Place a(n) balls in an urn, of which b(n) = A011900(n) are red; draw 2 balls without replacement; 2*Probability(2 red balls) = Probability(2 balls); this is equivalent to the Pell equation A(n)^2-2*B(n)^2 = -1 with a(n) = (A(n)+1)/2; b(n) = (B(n)+1)/2; and the fundamental solution (7;5) and the solution (3;2) for the unit form. - Paul Weisenhorn, Aug 03 2010

FORMULA

a(n) = (-1+sqrt(1+8*b(n)*(b(n)+1)))/2 with b(n) = A011900(n). [corrected by Michel Marcus, Dec 23 2022]

b(n+1) = round((2+(10+7*sqrt(2))*(3+2*sqrt(2))^n)/4) = A011900(n+1).

T(a(n)) = A011900(n)^2 + A001109(n), where T(n) is the n-th triangular number. See also A001653. - Charlie Marion, Apr 25 2011

CROSSREFS

Cf. A001009, A001541, A001542, A011900, A046729, A053141, A084159, A084703, A156035.

A055997

Numbers n such that n(n - 1)/2 is a square.

+10
17

1, 2, 9, 50, 289, 1682, 9801, 57122, 332929, 1940450, 11309769, 65918162, 384199201, 2239277042, 13051463049, 76069501250, 443365544449, 2584123765442, 15061377048201, 87784138523762, 511643454094369, 2982076586042450, 17380816062160329, 101302819786919522

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FORMULA

a(n) = 2*A001653(m)*A011900(n-m-1) +A002315(m)*A001652(n-m-1) - A001108(m) with m<n; otherwise, a(n) = 2*A001653(m)*A011900(m-n) - A002315(m)*A046090(m-n) - A001108(m). See Link to Generalized Proof re Square Triangular Numbers. - Kenneth J Ramsey, Oct 13 2011